NCERT Solutions For Class 11 Maths Chapter 8

NCERT Solutions Class 11 Maths Binomial Theorem

Ncert Solutions For Class 11 Maths Chapter 8 PDF Download

Here we are providing for the students of class 11 with downloadable forms of the NCERT Solutions for Class 11 Maths Chapter 8 pdf, so students can learn better and more effectively. The NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem contains topics such as the statement and proof of the binomial theorem for positive integral indices, general and middle term in binomial expansion, pascal’s triangle and more. Work on Solutions of NCERT for Class 11 Maths Chapter 8 to help you to improve your performance for your class 11 exam as well as for competitive exam preparation.

NCERT Solutions Class 11 Maths Chapter 8 Exercises

[a + b]n    =    [ nC0 × an ]  +  [ nC1 × (an – 1) × b ]  +  [ nC2 × (a n – 2) × b2 ]  +  [ nC3 × (an – 3 )× b3 ] + . . . . . . . . . . . . . + [ nCn – 1 × a × (bn – 1) ]  +  [ nCn × bn ]

 

\(\Rightarrow\) Binomial Coefficient:

The coefficients nC0, nC1, nC2 . . . . . . . . . nCn occurring in the Binomial Theorem are known as Binomial coefficients

 

Some conclusions from Binomial Theorem:

 

(i) [x + y]n = [ nC0 × (xn) ]  +  [ nC1 × (xn – 1) × y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  +  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

 

(ii) [x – y]n = [ nC0 × (xn) ]  –  [ nC1 × (xn – 1) ×  y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  –  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +(-1)n [ nCn × yn ]

 

(iii) [1 – x]n = [ nC0 ]  –  [ nC1 . x ]  +  [ nC2 . x2 ]  –  [ nC3 . x3 ] + . . . . . . . . . . . . . . . . . . + (-1)n [ nCn . xn ]

 

(iv) (a + b)n = \(\sum_{r\;=\;0}^{n}\) nCr (a)n – r × br

 

NOTE:

 

(1.) nCr = \(\frac{n!}{r!(n-r)!}\) where, n is a non-negative integer and [0 r n]

(2.) nC0 = nCn = 1

(3.) There are total (n + 1) terms in the expansion of (a + b)n

 

 

                                                    Exercise 8.1

 

Q.1: Expand the Expression (1 – 3x)5

 

Sol.

By using Binomial Theorem:

Since, (1 – x)n = [ nC0 ]  +  [ nC1 × (-x) ]  +  [ nC2 × (-x)2 ]  +  [ nC3 × (-x3) ] + . . . . . . . . . . . . . . . . . . + [ nCn × (-x)n ]

Therefore, (1 – 3x)5 = [ 5C0 ] – [ 5C1 × (3x)  ] + [ 5C2 × (3x)2  ] – [ 5C3 × (3x)3  ] + [ 5C4 × (3x)4 ] – [ 5C5 × (3x)5 ]

\(\\\Rightarrow\) 1 – [ 5 × (3x) ] + [ 10 × (9x2) ] – [ 10 × (27x3) ] + [ 5 × (81x4) ] – [ 1 × (243x5) ]

\(\\\Rightarrow\) 1 – 15x + 90x2 – 270x3 + 405x4 – 243x5

Therefore, (1 – 3x)5 = 1 – 15x + 90x2 – 270x3 + 405x4 – 243x5

 

 

Q.2: Expand the Expression \(\left ( \frac{5}{x}-\frac{x}{5} \right )^{5}\)

 

Sol.

By using Binomial Theorem:

Since, [x – y]n = [ nC0 × (xn) ]  –  [ nC1 × (xn – 1) ×  y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  –  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +(-1)n [ nCn × yn ]

Therefore, \(\left ( \frac{5}{x}-\frac{x}{5} \right )^{5}:\)

[5C0 × \(\left ( \frac{5}{x} \right )^{5}\) ] – [5C1 × \(\left ( \frac{5}{x} \right )^{4}\) ×  \(\left ( \frac{x}{5} \right )^{1}\) ] + [5C2 ×  \(\left ( \frac{5}{x} \right )^{3}\) × \(\left ( \frac{x}{5} \right )^{2}\) ] – [ 5C3 × \(\left ( \frac{5}{x} \right )^{2}\) × \(\left ( \frac{x}{5} \right )^{3}\) ] + [5C4 × \(\left ( \frac{5}{x} \right )^{1}\) × \(\left ( \frac{x}{5} \right )^{4}\) ] – [5C5 × \(\left ( \frac{x}{5} \right )^{5}]\\\)

\(\\\Rightarrow \left [ 1\times \frac{3125}{x^{5}}\right ]-\left [ 5\times \frac{625}{x^{4}}\times \frac{x}{5}\right ]+\left [ 10\times \frac{125}{x^{3}}\times \frac{x^{2}}{25}\right ]-\left [ 10\times \frac{25}{x^{2}}\times \frac{x^{3}}{125} \right ]+\left [ 5\times \frac{5}{x}\times \frac{x^{4}}{625} \right ]-\left [ 1\times \frac{x^{5}}{3125} \right ]\\\)

\(\\\Rightarrow \left [\frac{3125}{x^{5}}\right ]-\left [\frac{625}{x^{3}}\right ]+\left [ \frac{50}{x}\right ]-2x+\left [ \frac{3x^{3}}{25} \right ]-\left [\frac{x^{5}}{3125} \right ]\\\)

Therefore, \(\\\left ( \frac{5}{x}-\frac{x}{5} \right )^{5}\):

\(\\\left [\frac{x^{5}}{3125} \right ] +\left [ \frac{3x^{3}}{25} \right ]-2x+\left [ \frac{50}{x}\right ]-\left [\frac{625}{x^{3}}\right ]+ \left [\frac{3125}{x^{5}}\right ]\)

 

 

Q.3: Expand the Expression (3x – 2)6

 

Sol.

By using Binomial Theorem:

[x – y]n = [ nC0 × (xn) ]  –  [ nC1 × (xn – 1) ×  y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  –  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +(-1)n [ nCn × yn ]

Therefore, (3x – 2)6  = [ 6C0 × (3x)6 ]  –  [ 6C1 × (3x)5 × (2) ]  +  [ 6C2 × (3x)4 × (2)2 ]  –  [ 6C3 × (3x)3 × (2)3 ]  +  [ 6C4 × (3x)2 × (2)4 ]  –  [ 6C5 × (3x)1 × (2)5 ]  +  [ 6C6 ×  (2)6 ]

\(\\\Rightarrow\) [1 × (729x6)] – [6 × (243x5) × 2] + [15 × (81x4) × 4] – [20 × (27x3) × 8 ] + [15 × (9x2) × 16] – [6 × (3x) × 32] + [1 × 64]

\(\\\Rightarrow\) 729x6 – 2916x5 + 4860x4 – 4320x3 + 2160x2 – 576x + 64

Therefore, (3x – 2)6 = 729x6 – 2916x5 + 4860x4 – 4320x3 + 2160x2 – 576x + 64

 

 

Q.4: Expand the Expression \(\left ( \frac{4}{x}+\frac{x}{3} \right )^{5}\)

 

Sol.

By using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ]  +  [ nC1 × (xn – 1) × y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  –  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

Therefore, \(\left ( \frac{4}{x}+\frac{x}{3} \right )^{5}\):

[ 5C0 × \(\left ( \frac{4}{x} \right )^{5}\) ]  +  [ 5C1 × \(\left ( \frac{4}{x} \right )^{4}\) ×  \(\left ( \frac{x}{3} \right )^{1}\) ]  +  [ 5C2 ×  \(\left ( \frac{4}{x} \right )^{3}\) × \(\left ( \frac{x}{3} \right )^{2}\) ]  +  [ 5C3 × \(\left ( \frac{4}{x} \right )^{2}\) × \(\left ( \frac{x}{3} \right )^{3}\) ]  +  [ 5C4 × \(\left ( \frac{4}{x} \right )^{1}\) × \(\left ( \frac{x}{3} \right )^{4}\) ]  +  [ 5C5 × \(\left ( \frac{x}{3} \right )^{5}]\\\)

\(\\\Rightarrow \left [ 1\times \frac{1024}{x^{5}}\;\right ]+\left [ 5\times \frac{256}{x^{4}}\times \frac{x}{3}\;\right ]+\left [ 10\times \frac{64}{x^{3}}\times \frac{x^{2}}{9}\;\right ]+\left [ 10\times \frac{16}{x^{2}}\times \frac{x^{3}}{27} \right ]+\left [ 5\times \frac{4}{x}\times \frac{x^{4}}{81}\; \right ]+\left [ 1\times \frac{x^{5}}{243} \right ]\\\\\)

\(\\\Rightarrow \left [\frac{1024}{x^{5}}\right ]+\left [\frac{1280}{3x^{3}}\right ]+\left [ \frac{640}{9x}\right ]+\left [\frac{160x}{27}\right ]+\left [ \frac{20x^{3}}{81} \right ]+\left [\frac{x^{5}}{243} \right ]\\\)

Therefore, \(\left ( \frac{4}{x}+\frac{x}{3} \right )^{5}:\)

\(\\\Rightarrow \left [\frac{x^{5}}{243} \right ]+\left [ \frac{20x^{3}}{81} \right ]+\left [\frac{160x}{27}\right ]+\left [ \frac{640}{9x}\right ]+\left [\frac{1280}{3x^{3}}\right ]+\left [\frac{1024}{x^{5}}\right ]\)

 

 

Q.5: Expand the Expression \(\left ( \frac{1}{x}-\frac{x}{2} \right )^{6}\)

 

Sol.

By using Binomial Theorem:

Since, [x – y]n = [ nC0 × (xn) ]  –  [ nC1 × (xn – 1) ×  y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  –  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +(-1)n [ nCn × yn ]

Therefore, \(\left ( \frac{1}{x}-\frac{x}{2} \right )^{6}\):

[ 6C0 × \(\left ( \frac{1}{x} \right )^{6}\) ]  –  [ 6C1 × \(\left ( \frac{1}{x} \right )^{5}\) ×  \(\left ( \frac{x}{2} \right )^{1}\) ]  +  [ 6C2 ×  \(\left ( \frac{1}{x} \right )^{4}\) × \(\left ( \frac{x}{2} \right )^{2}\) ]  –  [ 6C3 × \(\left ( \frac{1}{x} \right )^{3}\) × \(\left ( \frac{x}{2} \right )^{3}\) ]  +  [ 6C4 × \(\left ( \frac{1}{x} \right )^{2}\) × \(\left ( \frac{x}{2} \right )^{4}\) ]  –  [ 6C5 × \(\left ( \frac{1}{x} \right )^{1}\) × \(\left ( \frac{x}{2} \right )^{5}\) ]  +  [ 6C6 ×\(\left ( \frac{x}{2} \right )^{6}]\\\)

\(\\\Rightarrow \left [ 1\times \frac{1}{x^{6}}\;\right ]-\left [ 6\times \frac{1}{x^{5}}\times \frac{x}{2}\;\right ]+\left [ 15\times \frac{1}{x^{4}}\times \frac{x^{2}}{4}\;\right ]-\left [ 20\times \frac{1}{x^{3}}\times \frac{x^{3}}{8} \right ]+\left [ 15\times \frac{1}{x^{2}}\times \frac{x^{4}}{16}\; \right ]-\left [ 6\times \frac{1}{x}\times \frac{x^{5}}{32} \right ]+\left [ 1\times \frac{x^{6}}{64} \right ]\\\)

\(\\\Rightarrow \left [ \frac{1}{x^{6}}\;\right ]-\left [ \frac{3}{x^{4}}\right ]+\left [ \frac{15}{4x^{2}}\right ]-\left [ \frac{5}{2} \right ]+\left [\frac{15x^{2}}{16}\; \right ]-\left [\frac{3x^{4}}{16} \right ]+\left [\frac{x^{6}}{64} \right ]\\\)

Therefore, \(\left ( \frac{1}{x}-\frac{x}{2} \right )^{6}:\)

\(\\\Rightarrow \left [\frac{x^{6}}{64} \right ]-\left [\frac{3x^{4}}{16} \right ]+\left [\frac{15x^{2}}{16} \right ]-\left [ \frac{5}{2} \right ]+\left [ \frac{15}{4x^{2}}\right ]-\left [ \frac{3}{x^{4}}\right ]+\left [ \frac{1}{x^{6}}\right ]\)

 

 

Q.6: By using Binomial Theorem, Evaluate (98)4

 

Sol.

(98)4 = (100 – 2)4

Now, by using Binomial Theorem:

Since, [x – y]n = [ nC0 × (xn) ]  –  [ nC1 × (xn – 1) ×  y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  –  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +(-1)n [ nCn × yn ]

Therefore,   (100 – 2)4 = [ 4C0 × (100)4 ]  –  [ 4C1 × (100)3 × (2) ]  +  [ 4C2 × (100)2 × (2)2 ]  –  [ 4C3 × 100 × (2)3 ]  +  [4C4 × (2)4]

\(\\\Rightarrow\) (1 × 100000000) – (4 × 1000000 × 2) + (6 × 10000 × 4) – (4 × 100 × 8 ) + (1 × 16)

\(\\\Rightarrow\) 100000000 – 8000000 + 240000 – 3200 + 16

Therefore, (98)4 = (100 – 2)4 = 92236816 = 9.2236816 × 107

 

 

Q.7: By using Binomial Theorem, Evaluate (105)5

 

Sol.

(105)5 = (100 + 5)5

Now, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ]  +  [ nC1 × (xn – 1) ×  y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  +  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

Therefore, (100 + 5)5 = [ 5C0 × (100)5 ]  +  [ 5C1 × (100)4 × (5) ]  +  [ 5C2 × (100)3 × (5)2 ]  +  [ 5C3 × (100)2 × (5)3 ]  +  [ 5C4 × (100) × (5)4 ]  +  [ 5C5 × (5)5]

\(\\\Rightarrow\) (1 × 10000000000)  +  (5 × 100000000 × 5) + (10 × 1000000 × 52 ) + (10 × 10000 × 53 ) + (5 × 100 × 54 ) + ( 1 × 55 )

\(\\\Rightarrow\) 10000000000 + 2500000000 + 250000000 + 12500000 + 312500 + 3125

Therefore, (105)5 = (100 + 5)5 = 12762815630 = 1.276281563 × 1010

 

 

Q.8: By using Binomial Theorem, Evaluate (104)4

 

Sol.

(104)4 = (100 + 4)4

Now, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ]  +  [ nC1 × (xn – 1) × y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  –  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

Therefore, (100 + 4)4 = [4C0 × (100)4 ] + [ 4C1 × (100)3 × (4) ] + [4C2 × (100)2 × (4)2] + [4C3 × 100 × (4)3] + [4C4 × (4)4 ]

\(\\\Rightarrow\) (1 × 100000000) + (4 × 1000000 × 4) + (6 × 10000 × 16) + (4 × 100 × 64 ) + (1 × 256)

\(\\\Rightarrow\) 100000000 + 16000000 + 960000 + 25600 + 256

Therefore, (104)4 = (100 + 4)4 = 116985856 = 1.16985856 × 108

 

 

Q.9: By using Binomial Theorem, Evaluate (95)5

 

Sol.

(95)5 = (100 – 5)5

Now, by using Binomial Theorem:

Since, [x – y]n = [ nC0 × (xn) ]  –  [ nC1 × (xn – 1) ×  y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  –  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +  (-1)n [ nCn × yn ]

Therefore, (100 – 5)5 = [ 5C0 × (100)5 ]  –  [ 5C1 × (100)4 × (5) ]  +  [ 5C2 × (100)3 × (5)2 ]  –  [ 5C3 × (100)2 × (5)3 ]  +  [ 5C4 × (100) × (5)4 ]  –  [ 5C5 × (5)5 ]

\(\\\Rightarrow\) (1 × 10000000000)  –  (5 × 100000000 × 5) + (10 × 1000000 × 52 ) – (10 × 10000 × 53 ) + (5 × 100 × 54 ) – ( 1 × 55 )

\(\\\Rightarrow\) 10000000000 – 2500000000 + 250000000 – 12500000 + 312500 – 3125

Therefore, (95)5 = (100 – 5)5 = 7737809375 = 7.737809375 × 109

 

 

Q.10: By using Binomial Theorem, determine which number is greater (1.1)10000 or 1000.

 

Sol.

Now, (1.1)10000 = (1 + 0.1)10000

Now, on applying Binomial Theorem, the first few terms in the expansion of (1 + 0.1)10000 are:

Since, [x + y]n = [ nC0 × (xn) ]  +  [ nC1 × (xn – 1) × y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  –  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

Therefore, (1 + 0.1)10000 = [ 10000C0 × (1)10000 ]  + [ 10000C1 × (1)9999 × 0.1 ] + other positive terms

\(\\\Rightarrow\)   [1 × 1] + [10000 × 1 × 0.1] + other positive terms

\(\\\Rightarrow\)   1 + 1000 + other positive terms > 1000

Therefore, (1.1)10000 is greater than 1000.

 

 

Q.11:  Find (a + b)5 – (a – b)5 . Hence evaluate \(\left(\sqrt{5}+\sqrt{3} \right )^{5}-\left ( \sqrt{5}-\sqrt{3}\right)^{5}\)

 

Sol.

Now, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ]  +  [ nC1 × (xn – 1) × y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  +  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

And, [x – y]n = [ nC0 × (xn) ]  –  [ nC1 × (xn – 1) ×  y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  –  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +  (-1)n [ nCn × yn ]

Therefore, (a + b)5= [ 5C0 × (a5) ]  +  [ 5C1 × (a4) × b ]  +  [ 5C2 ×  (a3) ×  b2 ]  +  [ 5C3 × (a2) × b3 ] + [ 5C4 × (a1) × b4 ] + [ 5C5 × b5 ]

And, [a – b]5 = [ 5C0 × (a5) ]  –  [ 5C1 × (a4) ×  b ]  +  [ 5C2 ×  (a3) ×  b2 ]  –  [ 5C3 × (a2) × b3 ] + [ 5C4 × (a1) × b4 ]  –  [ 5C5 × b5 ]

Therefore, (a + b)5 – (a – b)5 = [ 5Ca5  +  5C1 a4 b  +  5C2 a3b2  +  5C3 a2b3  +  5C4 ab4  +   5C5 b5 ]  –  [ 5C0 a5  –  5C1 a4b  +  5C2 a3b2  –  5C3 a2b3  5C4 ab4  –  5C5 b5 ]

\(\\\Rightarrow\)  [ 5Ca5  +  5C1 a4 b  +  5C2 a3b2  +  5C3 a2b3  +  5C4 ab4  +   5C5 b5   –   5C0 a5  +  5C1 a4b  –  5C2 a3b2  +  5C3 a2b3  –  5C4 ab4  +  5C5 b5 ]

\(\\\Rightarrow\)  2[ 5C1 a4b + 5C3 a2b3 + 5C5 b5 ] = 2[ 5a4b + 10a2b3 + b5 ]

Therefore, (a + b)5 – (a – b)5 = 2b[ 5a4 + 10a2 b2 + b4 ] . . . . . . (1)

Now, on substituting a = \(\sqrt{5}\) and b = \(\sqrt{3}\) in equation (1) we will get:

\(\\\left(\sqrt{5}+\sqrt{3} \right )^{5}-\left ( \sqrt{5}-\sqrt{3}\right)^{5} = 2\sqrt{3}\times \left [ 5( \sqrt{5})^{4} +10( \sqrt{5})^{2}\times ( \sqrt{3} )^{2}+(\sqrt{3})^{4}\right ]\\\)

\(\\\Rightarrow 2\sqrt{3}\;[125+150+9]=568\sqrt{3}\\\)

Therefore, \(\left(\sqrt{5}+\sqrt{3} \right )^{5}-\left ( \sqrt{5}-\sqrt{3}\right)^{5}\)=\(568\sqrt{3}\)

 

 

Q.12:  Find (1 + x)5 – (1 – x)5 . Hence evaluate \(\left ( 1+\sqrt{7} \right )^{5}-\left ( 1-\sqrt{7} \right )^{5}\)

 

Sol.

Now, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ]  +  [ nC1 × (xn – 1) × y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  +  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . + [ nCn × yn ]

And, [x – y]n = [ nC0 × (xn) ]  –  [ nC1 × (xn – 1) ×  y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  –  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . +  (-1)n [ nCn × yn ]

Therefore, (1 + x)5= [ 5C0 × (15) ]  +  [ 5C1 × (14) × (x) ]  +  [ 5C2 ×  (13) ×  (x)2 ]  +  [ 5C3 × (12) × (x)3 ] + [ 5C4 × (11) × (x)4 ] + [ 5C5 × (x)5 ]

And, [1 – x]5 = [ 5C0 × (15) ]  –  [ 5C1 × (14) × (x) ]  +  [ 5C2 ×  (13) × (x)2 ]  –  [ 5C3 × (12) × (x)3 ] + [ 5C4 × (11) × (x)4 ]  –  [ 5C5 × (x)5 ]

Therefore, (1 + x)5 – (1 – x)5 = [ 5C0  +  5C1 x  +  5C2 x2  +  5C3 x3  +  5C4 x4  +   5C5 x5 ]  –  [ 5C0  –  5C1 x  +  5C2 x2  –  5C3 x3  5C4 x4  –  5C5 x5 ]

\(\\\Rightarrow\)  [ 5C0  +  5C1 x  +  5C2 x2  +  5C3 x3  +  5C4 x4  +   5C5 x5   –  5C0  +  5C1 x  –  5C2 x2  +  5C3 x3  –  5C4 x4  +  5C5 x5 ]

\(\\\Rightarrow\)  2[ 5C1x  +  5C3 x3  +  5C5 x5 ] = 2[ 5x + 10x3 + x5 ]

Therefore, (1 + x)5 – (1 – x)5 = 2x[ 5 + 10x2 + x4 ] . . . . . . (1)

Now, on substituting x = \(\sqrt{7}\) in equation (1) we will get:

\(\\\Rightarrow \left ( 1+\sqrt{7} \right )^{7}-\left ( 1-\sqrt{7} \right )^{5}=2\sqrt{7}\left [ 5+10\left ( \sqrt{7} \right )^{2} + \left ( \sqrt{7} \right )^{4}\right ]\\\)

\(\\\Rightarrow 2\sqrt{7}\;[5+70+49]=248\sqrt{7}\\\)

Therefore, \(\left(1+\sqrt{7} \right )^{5}-\left ( 1-\sqrt{7}\right)^{5}\)=\(\;248\sqrt{7}\)

 

 

Q.13 Show that 7n+1 – 6n – 7 is divisible by 36, where n is the positive integer.

 

Sol.

Equation 7n+1 – 6n – 7 will be divisible by 36 if, 7n+1 – 6n – 7 = 36p [where p is any natural number]

Now, by Binomial Theorem:

Since, [1 + x]m = [ mC0 ]  +  [ mC1 . x ]  +  [ mC2 . x2 ]  +  [ mC3 . x3 ] + . . . . . . . . . . . . +  [ mCm . xm ]

Therefore, for x = 6 and m = n + 1 the equation becomes:

[1 + 6]n+1 = [ n+1C0 ]  +  [ n+1C1 × 6 ]  +  [ n+1C2 × (6)2 ]  +  [ n+1C3 × (6)3 ] + . . . . . . . . . . . . . + [ n+1Cn+1 × (6)n+1 ]

\(\\\Rightarrow\) [7]n + 1 = 1  +  [(n + 1) × 6]  +  62 [ n+1C2 n+1C3 × (6) + . . . . . . . . . + n+1Cn + 1 × (6)n–1 ]

\(\\\Rightarrow\) [7]n + 1 = 1  +  6n + 6  +  36 [ n+1C2  n+1C3 × (6) + . . . . . . . . . + n + 1Cn + 1 × (6)n – 1 ]

\(\\\Rightarrow\)   (7)n + 1 – 6 n – 7 = 36p

Where p is any natural number and p = [n + 1C2 +  n + 1C3 × (6) + . . . . . . . . . . . . . . . . . . + n + 1Cn + 1 × (6)n – 1

Therefore, 7n+1 – 6n – 7 is divisible by 36 [where n is a positive integer]

 

 

Q.14 Prove that \(\sum_{r\;=\;0}^{n}\) 2r × nCr = 3n

 

Sol.

By using Binomial Theorem:

Since, (a + b)n = \(\sum_{r\;=\;0}^{n}\) nCr (a)n – r × (b)r

Therefore, on substituting the value of a = 1 and b = 2 in the above equation we will get:

(1 + 2)n = \(\sum_{r\;=\;0}^{n}\) nCr (1)n – r × (2)r

\(\\\Rightarrow\) 3n = \(\sum_{r\;=\;0}^{n}\) nCr × (2)r

Therefore, \(\\\sum_{r\;=\;0}^{n}\) 2r × nCr = 3n

 

 

Q.15 Show that 5n – 4n always leaves remainder 5 when divided by 16, where n is the positive integer.

 

Sol.

Equation 5n – 4n will leave remainder 5 when divided by 16 if, 5n – 4n = 16p + 5 [where p is any natural number]

Now, by Binomial Theorem:

Since, [1 + x]n = [ nC0 ]  +  [ nC1 . x ]  +  [ nC2 . x2 ]  +  [ nC3 . x3 ] + . . . . . . . . . . . . . . . . . . +  [ nCn . xn ]

Therefore, for x = 4 the equation becomes:

[1 + 4]n = [ nC0 ]  +  [ nC1 × 4 ]  +  [ nC2 × (4)2 ]  +  [ nC3 × (4)3 ] + . . . . . . . . . . . . . . . . . . +  [ nCn × (4)n ]

\(\\\Rightarrow\) [5]n = 1  +  [(n + 1) × 4]  +  42 [ nC2 nC3 × (4) + . . . . . . . . . . . . . . . . . . + nCn × (4)n ]

\(\\\Rightarrow\) [5]n = 1  +  4n + 4  +  16 [ nC2  nC3 × (4) + . . . . . . . . . . . . . . . . . . + nCn × (4)n–1 ]

\(\\\Rightarrow\) 5n – 4n – 5 = 16p

Where p is any natural number and p = [ nC2 +  nC3 × (4) + . . . . . . . . . . . . . . . . . . + nCn × (4)n–1]

i.e.  5n – 4n = 16 + 5

Therefore, 5n – 4n will leave remainder 5 when divided by 16, where n is any natural number.

 

 

Q.16:  Find (a + b)6 – (a – b)6 . Hence evaluate: \(\left(\sqrt{2}+\sqrt{3} \right )^{6}-\left ( \sqrt{2}-\sqrt{3}\right)^{6}\)

 

Sol.

Now, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ]  +  [ nC1 × (xn – 1) × y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  +  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

And, [x – y]n = [ nC0 × (xn) ]  –  [ nC1 × (xn – 1) ×  y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  –  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +  (-1)n [ nCn × yn ]

Therefore, [a + b]6= [ 6C0 × (a6) ]  +  [ 6C1 × (a5) × b ]  +  [ 6C2 ×  (a4) ×  b2 ]  +  [ 6C3 × (a3) × b3 ] + [ 6C4 × (a2) × b4 ]  + [ 6C5 × a × b5 ]  +  [6C6 × b6]

And, [a – b]6 = [ 6C0 × (a6) ]  –  [ 6C1 × (a5) × b ]  +  [ 6C2 ×  (a4) ×  b2 ]  –  [ 6C3 × (a3) × b3 ] + [ 6C4 × (a2) × b4 ]  – [ 6C5 × a × b5 ]  +  [6C6 × b6]

Therefore, (a + b)6 – (a – b)6 = [ 6C0 a6  +  6C1 a5b  +  6C2 a4b6C3 a3b3  +  6C4 a2b4  +  6C5 ab5  +  6C6 b6] – [ 6C0 a6  – 6C1 a5b  +  6C2 a4b2  –  6C3 a3b6C4 a2b4  –  6C5 ab5  +  6C6 b6 ]

\(\\\Rightarrow\)  [ 6C0 a6  +  6C1 a5b  +  6C2 a4b6C3 a3b3  +  6C4 a2b4  +  6C5 ab5  +  6C6 b6  –  6C0 a6  + 6C1 a5b  –  6C2 a4b2  +  6C3 a3b–  6C4 a2b4  +  6C5 ab5  –  6C6 b6 ]

\(\\\Rightarrow\)  2[ 6C1 a5b + 6C3 a3b3 + 6C5 ab5 ] = 2[ 6a5b + 20a3b3 + 6ab5 ]

Therefore, (a + b)6 – (a – b)6 = 4ab[ 3a4 + 10a2 b2 + 3b4 ] . . . (1)

Now, on substituting a = \(\sqrt{2}\) and b = \(\sqrt{3}\) in equation (1) we will get:

\(\\(\sqrt{2}+\sqrt{3} )^{6}-(\sqrt{2}-\sqrt{3})^{6}=4  (\sqrt{2}. \sqrt{3})\left [ 3( \sqrt{2})^{4} +10 ( \sqrt{2})^{2}\times( \sqrt{3} )^{2}+3(\sqrt{3})^{4} \right ]\\\)

\(\\\Rightarrow 4\sqrt{6}\;[12+60+27]=396\sqrt{6}\\\)

Therefore, \(\left(\sqrt{2}+\sqrt{3} \right )^{6}-\left ( \sqrt{2}-\sqrt{3}\right)^{6}\)=\(\;396\sqrt{6}\)

 

 

Q.17: By using Binomial Theorem, Evaluate (91)4

 

Sol.

(91)4 = (100 – 9)4

Now, by using Binomial Theorem:

Since, [x – y]n = [ nC0 × (xn) ]  –  [ nC1 × (xn – 1) ×  y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  –  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +(-1)n [ nCn × yn ]

Therefore,   (100 – 9)4 = [ 4C0 × (100)4 ]  –  [ 4C1 × (100)3 × (9) ]  +  [ 4C2 × (100)2 × (9)2 ]  –  [ 4C3 × 100 × (9)3 ]  +  [4C4 × (9)4]

\(\\\Rightarrow\) (1 × 100000000) – (4 × 1000000 × 9) + (6 × 10000 × 81) – (4 × 100 × 729) + (1 × 6561)

\(\\\Rightarrow\) 100000000 – 36000000 + 4860000 – 291600 + 6561

Therefore, (91)4 = (100 – 9)4 = 68574961 = 6.8574961 × 107

 

 

Q.18: By using Binomial Theorem, Evaluate (107)5

 

Sol.

(107)5 = (100 + 7)5

Now, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ]  +  [ nC1 × (xn – 1) ×  y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  +  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

Therefore, (100 + 7)5 = [ 5C0 × (100)5 ]  +  [ 5C1 × (100)4 × (7) ]  +  [ 5C2 × (100)3 × (7)2 ]  +  [ 5C3 × (100)2 × (7)3 ]  +  [ 5C4 × (100) × (7)4 ]  +  [ 5C5 × (7)5]

\(\\\Rightarrow\) (1 × 10000000000)  +  (5 × 100000000 × 7) + (10 × 1000000 × 72 ) + (10 × 10000 × 73 ) + (5 × 100 × 74 ) + ( 1 × 75 )

\(\\\Rightarrow\) 10000000000 + 3500000000 + 490000000 + 34300000 + 1200500 + 16807

Therefore, (107)5 = (100 + 7)5 = 1402551731 = 1.402551731 × 1010

 

 

Q.19: Expand the Expression \(\left ( \frac{3}{2x}-\frac{x}{5} \right )^{6}\)

 

Sol.

By using Binomial Theorem:

Since, [x – y]n = [ nC0 × (xn) ]  –  [ nC1 × (xn – 1) ×  y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  –  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +(-1)n [ nCn × yn ]

Therefore, \(\\\left ( \frac{3}{2x}-\frac{x}{5} \right )^{6}:\\\)

[6C0 × \(\left ( \frac{3}{2x} \right )^{6}\) ]  –  [6C1 × \(\left ( \frac{3}{2x} \right )^{5}\) ×  \(\left ( \frac{x}{5} \right )^{1}\) ]  +  [6C2 ×  \(\left ( \frac{3}{2x} \right )^{4}\) × \(\left ( \frac{x}{5} \right )^{2}\) ]  –  [6C3 × \(\left ( \frac{3}{2x} \right )^{3}\) × \(\left ( \frac{x}{5} \right )^{3}\) ]  +  [6C4 × \(\left ( \frac{3}{2x} \right )^{2}\) × \(\left ( \frac{x}{5} \right )^{4}\) ]  –  [6C5 × \(\left ( \frac{3}{2x} \right )^{1}\) × \(\left ( \frac{x}{5} \right )^{5}\) ]  +  [6C6 ×\(\left ( \frac{x}{5} \right )^{6}]\\\)

\(\\\Rightarrow \left [ 1\times \frac{729}{64\times x^{6}}\;\right ]-\left [ 6\times \frac{243}{32\times x^{5}}\times \frac{x}{5}\;\right ]+\left [ 15\times \frac{81}{16\times x^{4}}\times \frac{x^{2}}{25}\;\right ]-\left [ 20\times \frac{27}{8x^{3}}\times \frac{x^{3}}{125} \right ]+\left [ 15\times \frac{9}{4x^{2}}\times \frac{x^{4}}{625}\; \right ]-\left [ 6\times \frac{3}{2x}\times \frac{x^{5}}{3125} \right ]+\left [ 1\times \frac{x^{6}}{15625} \right ]\\\)

\(\\\Rightarrow \left [\frac{729}{64\; x^{6}}\;\right ]-\left [\frac{729}{80\;x^{4}}\;\right ]+\left [\frac{243}{80\;x^{2}}\;\right ]-\left [\frac{27}{50}\right ]+\left [\frac{27x^{2}}{500}\;\right ]-\left [\frac{9x^{4}}{3125} \right ]+\left [\frac{x^{6}}{15625} \right ]\\\)

Therefore, \(\left ( \frac{3}{2x}-\frac{x}{5} \right )^{6}:\\\)

\(\Rightarrow \left [\frac{x^{6}}{15625} \right ]-\left [\frac{9x^{4}}{3125} \right ]+\left [\frac{27x^{2}}{500}\;\right ]-\left [\frac{27}{50}\right ]+\left [\frac{243}{80\;x^{2}}\;\right ]-\left [\frac{729}{80\;x^{4}}\;\right ]+\left [\frac{729}{64\; x^{6}}\;\right ]\)

 

 

Q.20: Expand the Expression (5x – 3)6

 

Sol.

By using Binomial Theorem:

Since, [x – y]n = [ nC0 × (xn) ]  –  [ nC1 × (xn – 1) ×  y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  –  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +(-1)n [ nCn × yn ]

Therefore, (5x – 3)= [ 6C0 × (5x)6 ]  –  [ 6C1 × (5x)5 × (3) ]  +  [ 6C2 × (5x)4 × (3)2 ]  –  [ 6C3 × (5x)3 × (3)3 ]  +  [ 6C4 × (5x)2 × (3)4 ]  –  [ 6C5 × (5x)1 × (3)5 ]  +  [ 6C6 ×  (3)6 ]

\(\\\Rightarrow\) [1 × (15625 x6)] – [6 × (3125 x5) × 3] + [15 × (625 x4) × 9] – [20 × (125 x3) × 27 ] + [15 × (25 x2) × 81] – [6 × (5x) × 243] + [1 × 729]

\(\\\Rightarrow\) 15625 x6 – 56250 x5 + 84375 x4 – 67500 x3 + 30375 x2 – 7290 x + 729

Therefore, (5x – 3)6 = 15625 x6 – 56250 x5 + 84375 x4 – 67500 x3 + 30375 x2 – 7290 x + 729

 

 

Exercise 8.2

 

Formulas:

 

1. The general term in the expansion of (a + b)n :

      Tr + 1 = nCr × (a)n – r × br

 

2. The middle term in the expansion of (a + b)n :

    (a).  If n is even:

        The middle term = \(\left ( \frac{n}{2}+1 \right )^{th}term\)

 

    (b).   If n is odd:

The middle term = \(\left ( \frac{n+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{n+1}{2} +1 \right )^{th}term\)    

 

 

Q.1: Find the Coefficient of x6 in the expansion of (x + 2)9

 

Sol.

Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Therefore, on substituting n = 9, a = x and b = 2 in the above expression we will get:

Tr + 1 = 9Cr × (x)9 – r × 2r . . . . . . . (1)

Now, on comparing the coefficient of x in equation (1) with x6, we will get:

i.e.         (x)9 – r = x6                                    

(or)         9 – r = 6

Therefore, r = 3

Now, on substituting the value of r in equation (1) we will get:

T4 = 9C3 × (x)9 – 3 × 23

\(\\\\\Rightarrow T_{4}=\frac{9!}{3!\;6!}\times x^{6}\times 8\\\\\)

\(\\\\\Rightarrow T_{4}=\frac{9\times 8\times 7\times 6!}{3\times 2\times 1\times 6!}\times 8\;x^{6}= 672\;x^{6}\\\\\)

Therefore, the Coefficient of x6 in the expansion of (x + 2)9 = 672

 

 

Q.2: Find the Coefficient of a6 b8 in the expansion of (2a – 3b)14

 

Sol.

Since, The general term in the expansion of (x + y)n is given by: Tr + 1 = nCr × (x)n – r × yr

Therefore, on substituting n = 14, x = 2a   and y = (-3b) in the above expression we will get:

Tr + 1 = 14Cr × (2a)14 – r × (-3b)r

i.e.    Tr + 1 = [14Cr × (2)14 – r (-3)r ] (a)14 – r × (b)r  . . . . . . . . . .  (1)

Now, on comparing the coefficients of a and b in equation (1) with a6 b8, we will get:

i.e.         a6 b8 = a14 – r br

Therefore,        r = 8

Now, on substituting the value of ‘r’ in equation (1) we will get:

T9 = [ 14C8 × (2)14 – 8 (-3)8 ] (a)14 – 8 × (b)8

i.e.     T9 = [ 14C8 × (2)6 (-3)8 ] a6 b8

\(\\\\\Rightarrow T_{9}=\frac{14!}{8!\;6!}\times 64\times 729\times a^{6}\;b^{8}\\\\\)

\(\\\\\Rightarrow T_{9}=\frac{14\times 13\times 12\times 11\times 10\times 9\times 8! }{6\times 5\times 4\times 3\times 2\times 1\times 8!}\times 64\times 729\times a^{6}\;b^{8}\\\\\)

\(\\\\\Rightarrow T_{9} = 3003\times 64\times 729\times a^{6}\;b^{8}=140107968\;\;a^{6}b^{8}\\\\\)

Therefore, the Coefficient of a6 b8 in the expansion of (2a – 3b)14 = 140107968

 

 

Q.3: Write the general term in the expansion of (x3 – y2)5

 

Sol.

The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Now, on substituting n = 5, a = x3 and b = y2 we will get:

Tr+1 = 5Cr × (x3)5 – r × (y2)r

Therefore, the general term in the expansion of (x3 – y2)5:

Tr+1 = 5Cr × (x)15 – 3 r × (y)2 r

 

 

Q.4: Write the general term in the expansion of (x4 – xy2)9

 

Sol.

The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Now, on substituting n = 9, a = x4 and b = xy2 we will get:

Tr+1 = 9Cr × (x4)9 – r × (xy2)r

\(\Rightarrow\) Tr+1 = 9Cr × (x)36 – 4 r × (x)2 r × (y)2 r

\(\Rightarrow\) Tr+1 = 9Cr × (x)36 – 4 r + 2 r × (y)2 r

\(\Rightarrow\) Tr+1 = 9Cr × (x)36 – 2 r × (y)2 r

Therefore, the general term in the expansion of (x4 – xy2)9:

Tr+1 = 9Cr × (x)36 – 2 r × (y)2r

 

 

Q.5: Find the 4th term in the expansion of (2x + 3y)10       

 

Sol.

Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Therefore, from the above equation the value of r should be 3; for finding out the values of 4th Term (T4)

Therefore, T3 + 1 = nC3 × (a)n – 3 × b3

Now, on substituting n =10, a = 2x and b = 3y we will get:

\(\\\Rightarrow\)   T3 + 1 = 10C3 × (2x)10 – 3 × (3y)3

\(\\\Rightarrow\)   T4 = 10C3 × (2x)7 × (3y)3

\(\\\\\Rightarrow T_{4}=\frac{10!}{3!\; 7!}\times 128\;x^{7}\times 27\;y^{3}\\\\\)

\(\\\\\Rightarrow T_{4}=\frac{10\times 9\times 8\times 7!}{3\times 2\times 1\times 7!}\times 3456\;x^{7}y^{3}\\\\\)

\(\\\\\Rightarrow T_{4} = 120\times 3456\;x^{7}y^{3}=414720\;x^{7}y^{3}\\\\\)

Therefore, 4th term in the expansion of (2x + 3y)10 = 414720 x7y3

 

 

Q.6 Find the 11th term in the expansion of \(\left ( 8x-\frac{1}{2\sqrt{x}} \right )^{15}\)

                                                                                                     

Sol.

Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Therefore, from the above equation the value of r should be 10 for finding out the values of 11th Term (T11)

Therefore, T10 + 1 = nC10 × (a)n – 10 × b10

Now, on substituting n =15, a = 8x and b = \(\left (-\frac{1}{2\sqrt{x}} \right )\) we will get:

\(\\\Rightarrow\)   T10 + 1 = 15C10 × (8x)15 – 10 × \(\left (-\frac{1}{2\sqrt{x}} \right )^{10}\\\)

\(\\\Rightarrow\)   T11 = 15C10 × (8x)5 × \(\left (-\frac{1}{2\sqrt{x}} \right )^{10}\\\)

\(\\\\\Rightarrow T_{11}=\frac{15!}{10!\; 5!}\times 8\times 8\times 8\times 8\times 8\times x^{5}\times \frac{1}{2^{10}}\times \left ( \frac{1}{\sqrt{x}} \right )^{10}\\\\\)

\(\\\\\\\Rightarrow T_{11}=\frac{15\times 14\times 13\times 12\times 11\times 10!}{5\times 4\times 3\times 2\times 1\times 10!}\times 32\times x^{5}\times \frac{1}{x^{5}}\\\\\)

\(\\\\\Rightarrow T_{11} = 3003\times 32\boldsymbol{=96096}\\\\\)

Therefore, 11th term in the expansion of \(\left ( 8x-\frac{1}{2\sqrt{x}} \right )^{15}\)= 96096

 

 

Q.7: In the expansion of \(\left ( 5-\frac{x^{4}}{10} \right )^{9}\), Find the middle terms.

 

Sol.

Here, n = 9

When n is odd then the middle terms in the expansion of (a + b)n are given by:

middle terms = \(\left ( \frac{n+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{n+1}{2} +1 \right )^{th}term\)

Therefore, the middle terms in the expansion of \(\left ( 5-\frac{x^{4}}{10} \right )^{9}\)are:

\(\Rightarrow \left ( \frac{9+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{9+1}{2} +1 \right )^{th}term\)

\(\Rightarrow\) 5th term and 6th term

Now, 5th and 6th terms in the expansion of \(\left ( 5-\frac{x^{4}}{10} \right )^{9}\) are:

T5 = T4 + 1 = 9C4 × (5)9 – 4 × \(\left ( -\frac{x^{4}}{10} \right )^{4}\)

\(\\\\\Rightarrow\)   T5 = 9C4 × (5)5 × \(\left ( -\frac{x^{4}}{10} \right )^{4}\\\\\)

\(\\\\\Rightarrow T_{5}=\frac{9!}{4!\; 5!}\times (5)^{5}\times \left ( -\frac{x^{4}}{10} \right )^{4}\\\\\)

\(\\\\\Rightarrow T_{5}=\frac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 1\times 5!}\times 5\times 5\times 5\times 5\times 5\times \frac{1}{10^{4}}\times x^{16}\\\\\)

\(\\\\\Rightarrow T_{5} = 126\times \frac{5}{16}\;x^{16} \boldsymbol{=\frac{315}{8}\;x^{16}}\\\\\)

Therefore, 5th term \(=\frac{315}{8}\;x^{16}\)

Now, T6 = T5 + 1 = 9C5 × (5)9 – 5 × \(\left ( -\frac{x^{4}}{10} \right )^{5}\)

\(\\\\\Rightarrow\)   T6 = 9C5 × (5)4 × \(\left ( -\frac{x^{4}}{10} \right )^{4}\\\\\)

\(\\\\\Rightarrow T_{6}=\frac{9!}{5!\; 4!}\times (5)^{4}\times \left ( -\frac{x^{4}}{10} \right )^{5}\\\\\)

\(\\\\\Rightarrow T_{6}=\frac{9\times 8\times 7\times 6\times 5!}{5!\times (4\times 3\times 2\times 1)}\times 5\times 5\times 5\times 5\times \frac{-1}{10^{5}}\times x^{20}\\\\\)

\(\\\\\Rightarrow T_{6} = -126\times \frac{1}{320}\;x^{20}\boldsymbol{=\frac{-63}{160}\;x^{20}}\\\\\)

Therefore, 6th term \(=\frac{63}{160}\;x^{20}\)

Therefore, 5th and 6th terms are the middle terms in the expansion of \(\left ( 5-\frac{x^{4}}{10} \right )^{9}\)

And also, 5th term \(=\frac{315}{8}\;x^{16}\) and 6th term \(=\frac{-63}{160}\;x^{20}\)

 

 

Q.8: In the expansion of \(\left ( \frac{x}{2}+8y \right )^{10}\), Find the middle term.

 

Sol.

Here, n = 10

When n is even, the middle term in the expansion of (a + b)n is given by:

\(\left ( \frac{n}{2}+1 \right )^{th}term\)

Therefore, the middle term in the expansion of \(\left ( \frac{x}{2}+8y \right )^{10}\) is:

\(\Rightarrow \left ( \frac{10}{2}+1 \right )^{th}term\) = 6th term

Now, 6th term in the expansion of \(\left ( \frac{x}{2}+8y \right )^{10}\) is:

T6 = T5 + 1 = 10C5 × \(\left ( \frac{x}{2} \right )^{10-5}\) × (8y)5

\(\\\Rightarrow\)   T6 = 10C5 × \(\left ( \frac{x}{2} \right )^{5}\) × (8y)5

\(\\\Rightarrow T_{6}=\frac{10!}{5!\; 5!}\times \left ( \frac{x}{2} \right )^{5}\times (8y)^{5}\\\)

\(\\\Rightarrow T_{6}=\frac{10\times 9\times 8\times 7\times 6\times 5!}{5\times 4\times 3\times 2\times 1\times 5!}\times \frac{x^{5}}{32}\times 8\times 8\times 8\times 8\times 8\times y^{5}\\\)

\(\\\Rightarrow T_{6} = 252\times 1024\;x^{5}y^{5}\boldsymbol{=258048\;x^{5}y^{5}}\\\)

Therefore, 6th term = 258048 x5 y5

Hence, the middle term in the expansion of \(\left ( \frac{x}{2}+8y \right )^{10}\) = 258048 x5 y5

 

 

Q.9: Prove that the coefficients of pand pb are equal in the expansion of (1 + p)a + b

 

Sol.

Since, The general term in the expansion of (x + y)n is given by: Tr + 1 = nCr × (x)n – r × (y)r

Now, on substituting n = (a + b), x =1 and y = p, we will get:

Tr + 1 = (a + b)Cr × (1)(a + b) – r × (p)r

\(\Rightarrow\) Tr + 1 = (a + b)Cr × (p)r . . . . . . . . . . . . (1)

Now, on comparing the coefficient of p in equation (1) with pa, we will get r = a

Therefore, from equation (1):

T a + 1 = (a + b)Ca × (p)a

\(\\\Rightarrow T_{a+1}=\frac{(a+b)!}{a!\;(a+b-a)!}\times p^{a}\\\)

\(\\\Rightarrow T_{a+1}=\frac{(a+b)!}{a!\;b!}\times p^{a}\\\)

Therefore, the coefficient of (p)a =  \(\frac{(a+b)!}{a!\;b!}\) . . . . . . . . . . (2)

Now, on comparing the coefficient of p in equation (1) with pb, we will get r = b

Therefore, from equation (1):

T b + 1 = (a + b)Cb × (p)b

\(\\\\\Rightarrow T_{b+1}=\frac{(a+b)!}{b!\;(a+b-b)!}\times p^{b}\\\\\)

\(\\\\\Rightarrow T_{b+1}=\frac{(a+b)!}{b!\;a!}\times p^{b}\\\\\)

Therefore, the coefficient of (p)b =  \(\frac{(a+b)!}{b!\;a!}\) . . . . . . . . . . (3)

Now, on comparing equation (2) and equation (3) we conclude that the coefficients of pand pb are equal in the expansion of (1 + p)a + b.

Hence, Proved

 

 

Q.10: The coefficients of the (k + 1)th, kth and (k – 1)th terms in the expansion of (x + 1)n are in the ratio 5 : 3 : 1. Find the values of n and k.

 

Sol.

Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Now on substituting a = x and b = 1 in the above equation:

Tr + 1 = nCr × (x)n – r × (1)r

Now, (k + 1)th term in the expansion of (x + 1)n :

T(k) + 1 = nCk × (x)n – k × 1(k )

i.e.  T(k + 1) = nCk × (x)n – k

\(\\\Rightarrow T_{k+1} = \frac{n!}{k!\;(n-k)!}\times (x)^{n-k}\\\)

Therefore the coefficient of T(k + 1)th term = \(\frac{n!}{k!\;(n-k)!}\\\\\)

Now, (k)th term in the expansion of (x + 1)n :

T(k – 1) + 1 = nCk – 1 × (x)n – (k – 1) × 1(k – 1)

i.e.  Tk = nCk – 1 × (x)n – k + 1

\(\\\\\Rightarrow T_{k} = \frac{n!}{(k-1)!\;(n-k+1)!}\times (x)^{n-k+1}\\\\\)

Therefore the coefficient of  T(k)th term = \(\frac{n!}{(k-1)!\;(n-k+1)!}\)

And, (k – 1)th term in the expansion of (x + 1)n :

T(k – 2 )+ 1 = nCk – 2 × (x)n – (k – 2) × 1(k – 2)

i.e.  Tk – 1 = nCk – 2 × (x)n – k + 2

\(\\\\\Rightarrow T_{k-1} = \frac{n!}{(k-2)!\;(n-k+2)!}\times (x)^{n-k+2}\\\\\)

Therefore the coefficient of  T(k – 1)th term = \(\frac{n!}{(k-2)!\;(n-k+2)!}\)

Now, according to the given conditions: Coefficients of T(k + 1), Tk and Tk – 1 are in the ratio of 5 : 3 : 1

Therefore, \(\frac{T_{k-1}}{T_{k}}=\frac{1}{3}\\\)

\(\\\\\Rightarrow \left [ \frac{n!}{(k-2)!\;(n-k+2)!} \right ]\div \left [ \frac{n!}{(k-1)!\;(n-k+1)!} \right ]=\frac{1}{3}\\\\\)

\(\\\\\Rightarrow \frac{(k-1)!\;(n-k+1)!}{(k-2)!\;(n-k+2)!}=\frac{1}{3}\\\\\)

\(\\\\\Rightarrow \frac{(k-1)(k-2)!\;(n-k+1)!}{(k-2)!\;(n-k+2)(n-k+1)!}=\frac{1}{3}\\\\\)

\(\\\Rightarrow \frac{k-1}{n-k+2}=\frac{1}{3}\\\)

\(\\\Rightarrow 3k-3=n-k+2\\\)

(or)              n – 4k + 5 = 0 . . . . . . . . . . . . . . . .  (1)

And,        \(\frac{T_{k+1}}{T_{k}}=\frac{5}{3}\\\)

\(\\\Rightarrow \left [ \frac{n!}{k!\;(n-k)!} \right ]\div \left [ \frac{n!}{(k-1)!\;(n-k+1)!} \right ]=\frac{5}{3}\\\)

\(\\\Rightarrow \frac{(k-1)!\;(n-k+1)!}{k!\;(n-k)!}=\frac{5}{3}\\\)

\(\\\Rightarrow \frac{(k-1)!\;[(n-k+1)\;(n-k)!]}{[k\;(k-1)!]\;(n-k)!}=\frac{5}{3}\\\)

\(\\\Rightarrow \frac{n-k+1}{k}=\frac{5}{3}\\\)

\(\\\Rightarrow 3n-3k+3=5k\\\)

(or) 3n – 8k + 3 = 0 . . . . . . . . . . . . . . . . . . (2)

On multiplying equation (1) by 3 and subtracting it to equation (2) we will get:

(3n – 8k + 3) – 3(n – 4k + 5) = 0

\(\\\Rightarrow\) 3n – 8k + 3 – 3n + 12k – 15 = 0

\(\\\Rightarrow\) 4k = 12

Therefore, k = 3

Now on substituting the value of k in equation (2) we will get:

3n – 8(3) + 3 = 0

Therefore, n = 7

 

 

Q.11: Prove that the coefficient of ap in the expansion of (1 + 3a)2p – 1 is half the coefficient of ap in the expansion of (1 + 3a)2p

 

Sol.

Since, The general term in the expansion of (x + y)n is given by: Tr + 1 = nCr × (x)n – r × (y)r

For (1 + 3a)2p:

On putting n = 2p, x =1 and y = 3a, we will get:

Tr + 1 = (2p)Cr × (1)(3p) – r × (3a)r

\(\\\\\Rightarrow\) Tr + 1 = (2p)Cr × (3a)r . . . . . . . . . . . . (1)

Now, on comparing the coefficient of a in equation (1) with ap, we will get r = p

Therefore, from equation (1):

T p + 1 = (2p)Cp × (3a)p

\(\\\Rightarrow T_{p+1}=\frac{(2p)!}{p!\;(2p-p)!}\times (3a)^{p}\\\\\)

\(\\\Rightarrow T_{p+1}=\frac{(2p)!}{p!\;p!}\times 3^{p}\times (a)^{p}\\\\\)

Therefore, the coefficient of ap \(=\frac{(2p)!}{p!\;p!}\times 3^{p}\) . . . . . . . . . . . . . . . . (2)

Now, for (1 + 3a)2p – 1 :

On substituting n = (2p – 1), x =1 and y = 3a, we will get:

Tr + 1 = (2p – 1)Cr × (1)(2p – 1) – r × (3a)r

\(\Rightarrow\) Tr + 1 = (2p – 1)Cr × (3a)r . . . . . . . . . . . . (1)

Now, on comparing the coefficient of ‘a’ in equation (1) with ‘ap ’, we will get r = p

Therefore, from equation (1):

T p + 1 = (2p – 1)Cp × (3a)p

\(\\\\\Rightarrow T_{p+1}=\frac{(2p-1)!}{p!\;(2p-1-p)!}\times (3a)^{p}\\\\\)

\(\\\\\Rightarrow T_{p+1}= \frac{(2p-1)!}{p!\;(p-1)!}\times (3)^{p}\times (a)^{p}\\\\\)

\(\\\\\Rightarrow T_{p+1}=\frac{\frac{(2p)!}{2p}}{p!\times \frac{p!}{p}}\times (3)^{p}.(a)^{p}=\frac{(2p)!}{2(p!)^{2}}\times 3^{p}.a^{p}\\\)

Therefore, the coefficient of ap = \(=\frac{(2p)!}{2(p!)^{2}}\times 3^{p}\) . . . . . . . . . . (3)

Now, on comparing equation (2) and equation (3):

\(\\\\\Rightarrow \frac{(2p)!}{p!\;p!}\times 3^{p}=\frac{(2p)!}{p!\;p!}\times 3^{p}\\\\\)

\(\\\Rightarrow\)  (2p)Cp × 3=  \(\frac{1}{2}\) (2p – 1)Cp × 3p

Therefore, the coefficient of ap in the expansion of (1 + 3a)2p – 1 is half the coefficient of ap in the expansion of (1 + 3a)2p

 

 

Q.12: For what values of ‘m’ the coefficient of x2 in the expansion of (1+ 2x)k is 140.

 

Sol.

Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × (b)r

Now, on substituting n = m, a =1 and b = 2x we will get:

Tr + 1 = mCr × (1)m – r × (2x)r

\(\Rightarrow\) Tr + 1 = mCr × (2)r × (x)r . . . . . . . . . . . . (1)

Now, on comparing the coefficient of x in equation (1) with x2, we will get r = 2

Therefore, from equation (1):

T 2 + 1 = mC2 × (2)2 × (x)2

Since, the coefficient of x2 = 140   [GIVEN]

Therefore,    140 = mC2 × (2)2

\(\Rightarrow 84=\left [ \frac{m!}{2!\;\;(m-2)!} \right ]\times 4\\\\\)

\(\Rightarrow 21=\frac{m\times (m-1)\;\;(m-2)!}{[2\times 1]\times (m-2)!}\\\\\)

\(\Rightarrow 42 = m^{2}-m\)

Therefore, m2 – m – 42 = 0

Now, by splitting of middle term method, the roots of this quadratic equation are:

\(\Rightarrow\)m2 – (7 – 6)m – 42 = 0

\(\Rightarrow\)m2 – 7m + 6m – 42 = 0

\(\Rightarrow\) m(m – 7) +6(m – 7) = 0

i.e.    (m + 6) (m – 7) = 0

Therefore, m = 7 or m = -6

Hence, for the coefficient of x2 in the expansion of (1+ 2x)k to be 140, the value of ‘m’ should be 7.

 

 

Q.13: In the expansion of (2x+3y)9, Find the middle terms.

 

Sol.

Here, n = 9

When n is odd then the middle terms in the expansion of (a + b)n are given by:

middle terms = \(\left ( \frac{n+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{n+1}{2} +1 \right )^{th}term\)

Therefore, the middle terms in the expansion of (2x+3y)9 are:

\(\Rightarrow \left ( \frac{9+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{9+1}{2} +1 \right )^{th}term\)

\(\Rightarrow\) 5th term and 6th term

Now, 5th and 6th terms in the expansion of (2x+3y)9 are:

T5 = T4 + 1 = 9C4 × (2x)9 – 4 × (3y)4

\(\Rightarrow\)   T5 = 9C4 × (2x)5 × (3y)4

\(\Rightarrow T_{5}=\frac{9!}{4!\; 5!}\times (2x)^{5}\times (3y)^{4}\)

\(\Rightarrow T_{5}=\frac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 1\times 5!}\times 32x^{5}\times 81y^{4}\\\\\)

\(\Rightarrow T_{5} = 126\times 2592\;x^{5}y^{4}\boldsymbol{=414720\;x^{5}y^{4}}\\\)

Therefore, 5th term = 414720 x5 y4

Now, T6 = T5 + 1 = 9C5 × (2x)9 – 5 × (3y)5

\(\Rightarrow\) T6 = 9C5 × (2x)4 × (3y)5

\(\Rightarrow T_{6}=\frac{9!}{5!\times 4!}\times (2x)^{4}\times (3y)^{5}\)

\(\Rightarrow T_{6}=\frac{9\times 8\times 7\times 6\times 5!}{5!\times (4\times 3\times 2\times 1)}\times 16x^{4}\times 243y^{5}\\\\\)

\(\Rightarrow T_{6} = 126\times 3888\;x^{4}y^{5}\boldsymbol{=489888\;x^{4}y^{5}}\\\)

Therefore, 6th term = 489888 x4 y5

Hence, 5th and 6th terms are the middle terms in the expansion of (2x+3y)9

And also, 5th term = 414720 x5 y4 and 6th term = 489888 x4 y5

 

 

Q.14: Find the Coefficient of x6 in expansion of (x + 3)11

 

Sol.

Since, The general term in the expansion of (a + b)is given by:  Tr + 1 = nCr × (a)n – r × br

Therefore, on substituting n = 11, a = x and b = 3 in the above expression we will get:

Tr + 1 = 11Cr × (x)11 – r × 3r . . . . . . . (1)

Now, on comparing the coefficient of x in equation (1) with x7, we will get:

i.e.         (x)11 – r = x6

(or)         11 – r = 6

Therefore, r = 5

Now, on substituting r in equation (1) we will get:

T5 + 1 = 11C5 × (x)11 – 5 × 35

\(\Rightarrow T_{6}=\frac{11!}{5!\times 6!}\times x^{6}\times 3^{5}\)

\(\Rightarrow T_{6}=\frac{11\times 10\times 9\times 8\times 7\times 6!}{5\times 4\times 3\times 2\times 1\times 6!}\times 243\;x^{6}\\\\\)

\(\Rightarrow T_{6} = 462\times 243\;x^{6}\boldsymbol{=112266\;x^{6}}\)

Therefore, the Coefficient of x6 in the expansion of (x + 3)11 = 112266

 

 

Q.15: Write the general term in the expansion of (x2y3 – x3 y2)7

 

Sol.

The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Now, on substituting n = 7, a = x2y3 and b = x3 y2 we will get:

Tr+1 = 7Cr × (x2y3)7 – r × (x3 y2)r

\( \Rightarrow\) Tr + 1 = 7Cr × (x2)7 – r × (y3)7 – r × (x)3 r × (y)2 r

\( \Rightarrow\) Tr + 1 = 7Cr × (x)14 – 2 r × (x)3 r× (y)2 r× (y)21 – 3r

\( \Rightarrow\) Tr + 1 = 7Cr × (x)14 – 2 r + 3r  ×  (y)21 – 3r + 2 r

i.e.  Tr + 1 = 7Cr × (x)14 + r  ×  (y)21 – r

Therefore, the general term in the expansion of (x2y3 – x3 y2)7:

Tr + 1 = 7Cr × (x)14 + r × (y)21 – r

 

 

Miscellaneous Exercise:

 

Q.1: If 1st term, 2nd term and 3rd term in the expansion of (a + b)n are 729, 7290 and 30375 respectively then find the value of ‘a’, ‘b’ and ‘c’.

 

Sol.

The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Now, according to the given conditions:

T(0 + 1) = nC0 × (a)n – 0 × b0

T1 = an

i.e.   an = 729 . . . . . . . . . . . . . . . . . . (1)

Also, T(1 + 1) = nC1 × (a)n – 1 × b1

T2 = nC1 × (a)n – 1 × b

i.e.   7290 = nC1 × (a)n – 1 × b

\(\\\Rightarrow 7290=\frac{n!}{1!\;\;(n-1)!}\times a^{n}\times a^{-1}\times b\\\\\)

\(\\\\\Rightarrow 7290=\frac{n(n-1)!}{(n-1)!}\times a^{n}\times \frac{b}{a}\;\;\;\)

Therefore, 7290 = n × an × \(\frac{b}{a}\) . . . . . . . . . . . . . . . . (2)

And, T(2 + 1) = nC2 × (a)n – 2 × b2

T3 = nC2 × (a)n – 2 × b2

i.e.   30375 = nC2 × (a)n – 2 × b2

\(\\\Rightarrow 30375=\frac{n!}{2!\;\;(n-2)!}\times a^{n}\times a^{-2}\times b^{2}\\\)

\(\\\Rightarrow 30375=\frac{n\;(n-1)\;(n-2)!}{(2\times 1)\;\;(n-2)!}\times a^{n}\times \left ( \frac{b}{a} \right )^{2}\\\)

Therefore, 60750 = n (n – 1) × an × \(\left ( \frac{b}{a} \right )^{2}\)  . . . . . . . . . . . . . . . . (3)

Now, on substituting equation (1) in equation (2) we will get:

\(\\\Rightarrow 7290 = n\times (729)\times \frac{b}{a}\\\)

\(\\\Rightarrow \frac{10a}{b}=n\\\)

Therefore, n = \(\frac{10a}{b}\) . . . . . . . . . . . . . . . . . (4)

Now, on dividing equation (2) and equation (3) we will get:

\(\\\Rightarrow \frac{n \times a^{n} \times \frac{b}{a}}{n\times (n-1)\times a^{n}\times \left ( \frac{b}{a} \right )^{2} }=\frac{7290}{60750}\\\)

\(\\\Rightarrow \frac{3}{25}=\frac{1}{(n-1)\times \frac{b}{a}}\\\)

\(\\\Rightarrow 3n-3=\frac{25a}{b}\\\)

Now, from equation (4):

\(\frac{a}{b}\) = \(\frac{n}{10}\)

\(\Rightarrow 3n-3=25\times \left ( \frac{n}{10} \right )\)

\(\Rightarrow 30n-30=25n\)

i.e.   30n – 30 = 25n

Therefore, n = 6

Now, on substituting the value of ‘n’ in equation (1) we will get:

\(\Rightarrow\) a6 = 729

i.e.  a6 = 36

Therefore, a = 3

Now, from equation (4):

\(\\\Rightarrow\) 6 = \(\frac{30}{b}\\\)

Therefore, b = 5

 

 

Q.2: If the coefficients of z2 and z3 are equal in the expansion of (3 + bz)9. Find the value of ‘b’

 

Sol.

Since, The general term in the expansion of (x + y)n is given by: Tr + 1 = nCr × (x)n – r × yr

Therefore, on substituting n = 9, x = 3 and y = bz in the above expression we will get:

Tr + 1 = 9Cr × (3)9 – r × (bz)r

Tr + 1 = 9Cr × (3)9 – r × br × zr . . . . . (1)

Now, on comparing the coefficient of z in equation (1) with z2, we will get: r = 2

On substituting the value of r in equation (1) we will get:

T(2 + 1) = 9C2 × (3)9 – 2 × b2 × z2

\(\\\Rightarrow T_{3}=\frac{9!}{2!\;7!}\times 3^{7}\times b^{2}\times z^{2}\\\)

\(\\\Rightarrow T_{3}=\frac{9\times 8\times 7!}{2\times 1\times 7!}\times 3^{7}\times b^{2}\times z^{2}\\\)

\(\\ \Rightarrow T_{3} = 36\times 3^{7}\times b^{2}\times z^{2}\) . . . . . . . . . . . . (2)

Now, on comparing the coefficient of z in equation (1) with z3, we will get: r = 3

On substituting the value of ‘r’ in equation (1) we will get:

T(3 + 1) = 9C3 × (3)9 – 3 × b3 × z3

\(\\\Rightarrow T_{4}=\frac{9!}{3!\;6!}\times 3^{6}\times b^{3}\times z^{3}\\\)

\(\\\Rightarrow T_{4}=\frac{9\times 8\times 7\times 6!}{3\times 2\times 1\times 6!}\times 3^{6}\times b^{3}\times z^{3}\\\)

\(\\\Rightarrow T_{4} = 84\times 3^{6}\times b^{3}\times z^{3} . . . . . . . . . (3)\)

Now, according to the given conditions:

Coefficient of z2 = Coefficient of z3

Therefore from equation (2) and equation (3):

\(\\\Rightarrow 36\times 3^{7}\times b^{2}=84\times 3^{6}\times b^{3}\\\)

\(\\\Rightarrow b = \frac{36\times 3^{7}}{84\times 3^{6}}\\\)

Therefore, the value of b = \(\frac{9}{7}\)

 

 

Q.3: Find the coefficient of x6 in the product of (1 + 3x)7 (1 – 2x)6 by using binomial theorem.

 

Sol.

By using Binomial Theorem:

Since, (1 + x)n = [ nC0 ]  +  [ nC1 × (x) ]  +  [ nC2 × (x)2 ]  +  [ nC3 × (x3) ] + . . . . . . . . . . . . . . . . . . + [ nCn × (x)n ]

Therefore, (1 + 3x)7 = [ 7C0 ]  +  [ 7C1 × (3x)  ]  +  [ 7C2 × (3x)]  +  [ 7C3 × (3x)]  + [ 7C4 × (3x)4 ]  +  [ 7C5 × (3x5) ]  +  [ 7C6 × (3x)6 ]  +  [ 7C7 × (3x)7 ]

\(\\\Rightarrow\) 1 + [ 7 × (3x) ] + [ 21 × (9x2) ] + [ 35 × (27x3) ] + [ 35 × (81x4) ] + [ 21 × (243x5) ] + [ 7 × 729x6 ] + [ 1 × 2187x7 ]

\(\\\Rightarrow\) 1 + 21x + 189x2 + 945x3 + 2835x4 + 5103x5 + 5103x6 + 2187x7

Therefore, (1 + 3x)7 = 1 + 21x + 189x2 + 945x3 + 2835x4 + 5103x5 + 5103x6 + 2187x7

Now, (1 – 2x)6:

By using Binomial Theorem:

(1 – 2x)6 = [ 6C0 ]  –  [ 6C1 × (2x)  ]  +  [ 6C2 × (2x)]  –  [ 6C3 × (2x)]  + [ 6C4 × (2x)4 ]  –  [ 6C5 × (2x5) ]  +  [ 6C6 × (2x)6 ]

\(\\\Rightarrow\) 1 – [ 6 × (2x) ] + [ 15 × (4x2) ] – [ 20 × (8x3) ] + [ 15 × (16x4) ] – [ 6 × (32x5) ] + [ 1 × 64x6 ]

\(\\\Rightarrow\) 1 – 12x + 60x2 – 160x3 + 240x4 – 192x5 + 64x6

Therefore, (1 – 2x)6 = 1 – 12x + 60x2 – 160x3 + 240x4 – 192x5 + 64x6

Now, (1 + 3x)7 (1 – 2x)6:

=(1 + 21x + 189x2 + 945x3 + 2835x4 + 5103x5 + 5103x6 + 2187x7) × (1 – 12x + 60x2 – 160x3 + 240x4 – 192x5 + 64x6)

Now, it not required to multiply the above equation, only terms with x5 are required and that can be identified by analyzing the above equation.

\(\\\Rightarrow\) [{1 × (- 192x5)} + {(21x) × (240x4)} + {(189x2) × (- 160x3)} + {(945x3) × (60x2)} + {(2835x4) × (- 12x)} + {(5103x5) × (1)}]

\(\\\Rightarrow\) [ -192x5 + 5040x5 – 30240x5+ 56700x5 – 34020x5 + 5103x5 ]

\(\\\Rightarrow\) 2391 x5

Therefore, the coefficient of x5 = 2391 

 

 

Q.4: Show that (a – b) is a factor of (an – bn), where ‘n’ is a positive integer and ‘a’ and ‘b’ are distinct integers.

 

Sol.

Now, for (a – b) to be factor of (an – bn), (an – bn) = p(a – b)    [where p is any natural number]

We can write an = (a – b + b)n = [(a – b) + b]n

Now, by Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ]  +  [ nC1 × (xn – 1) × y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  +  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

Therefore, for x = (a – b) and y = b the equation becomes:

[(a – b) + b]n = [ nC0 × (a – b)n ]  +  [ nC1 × (a – b)n – 1 × b ]  +  [ nC2 ×  (a – b)n – 2 ×  b2 ]  +  [ nC3 × (a – b)n – 3 × b3 ] + . . . . . . . . . . . . . + [ nCn – 1 ×  (a – b) ×  bn – 1 ]   + [ nCn × bn ]

\(\\\Rightarrow\) [a]n = [(a – b)n]  +  [ nC1 × (a – b)n – 1 × b ]  +  [ nC2 ×  (a – b)n – 2 ×  b2 ]  +  [ nC3 × (a – b)n – 3 × b3 ] + . . . . . . . . . . . . . + [ nCn – 1 ×  (a – b) ×  bn – 1 ]   +  [ bn ]

Now, an – bn = [(a – b)n] + [ nC1 × (a – b)n – 1 × b ] + [ nC2 ×  (a – b)n – 2 ×  b2 ] + [ nC3 × (a – b)n – 3 × b3 ] + . . . . . . . . . . . . . + [ nCn – 1 ×  (a – b) ×  bn – 1 ] + [ bn ] – [ bn ]

\(\\\Rightarrow\)   an – bn = (a – b) × [ (a – b)n – 1 + nC1 (a – b)n – 2 b + nC2 (a – b)n – 3 b2 + nC3 (a – b)n – 4 b3 + . . . . . . . . . . . . . + nCn – 1 bn – 1 ]

\(\\\Rightarrow\) (an – bn) = (a – b) × p

Where, p = [ (a – b)n – 1 + nC1 (a – b)n – 2 b + nC2 (a – b)n – 3 b2 + nC3 (a – b)n – 4 b3 + . . . . . . . . . . . . . + nCn – 1 bn – 1 ] is any natural number.

Therefore, (a – b) is a factor of (an – bn), where ‘n’ is a positive integer.

 

 

Q.5:  Evaluate \(\left(\sqrt{7}+\sqrt{5} \right )^{6}-\left ( \sqrt{7}-\sqrt{5}\right)^{6}\)

 

Sol.

Find (a + b)6 – (a – b)6

Now, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ]  +  [ nC1 × (xn – 1) × y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  +  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

And, [x – y]n = [ nC0 × (xn) ]  –  [ nC1 × (xn – 1) ×  y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  –  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +  (-1)n [ nCn × yn ]

Therefore, [a + b]6= [ 6C0 × (a6) ]  +  [ 6C1 × (a5) × b ]  +  [ 6C2 ×  (a4) ×  b2 ]  +  [ 6C3 × (a3) × b3 ] + [ 6C4 × (a2) × b4 ]  + [ 6C5 × a × b5 ]  +  [6C6 × b6]

And, [a – b]6 = [ 6C0 × (a6) ]  –  [ 6C1 × (a5) × b ]  +  [ 6C2 ×  (a4) ×  b2 ]  –  [ 6C3 × (a3) × b3 ] + [ 6C4 × (a2) × b4 ]  – [ 6C5 × a × b5 ]  +  [6C6 × b6]

Therefore, (a + b)6 – (a – b)6 = [ 6C0 a6  +  6C1 a5b  +  6C2 a4b6C3 a3b3  +  6C4 a2b4  +  6C5 ab5  +  6C6 b6] – [ 6C0 a6  – 6C1 a5b  +  6C2 a4b2  –  6C3 a3b6C4 a2b4  –  6C5 ab5  +  6C6 b6 ]

\(\\\Rightarrow\)  [ 6C0 a6  +  6C1 a5b  +  6C2 a4b6C3 a3b3  +  6C4 a2b4  +  6C5 ab5  +  6C6 b6  –  6C0 a6  + 6C1 a5b  –  6C2 a4b2  +  6C3 a3b–  6C4 a2b4  +  6C5 ab5  –  6C6 b6 ]

\(\\\Rightarrow\)  2[ 6C1 a5b + 6C3 a3b3 + 6C5 ab5 ] = 2[ 6a5b + 20a3b3 + 6ab5 ]

Therefore, (a + b)6 – (a – b)6 = 4ab[ 3a4 + 10a2 b2 + 3b4 ] . . . . . . (1)

Now, on substituting a = \(\sqrt{7}\) and b = \(\sqrt{5}\) in equation (1) we will get:

\(\\\left(\sqrt{7}+\sqrt{5} \right )^{6}-\left ( \sqrt{7}-\sqrt{5}\right)^{6}:\\\)

\(\\=4( \sqrt{7}\times \sqrt{5} )\left [ (3 \sqrt{7} )^{4} +10\times (\sqrt{7} )^{2}\times ( \sqrt{5})^{2}+3(\sqrt{5})^{4} \right ]\\\)

\(\\\Rightarrow 4\sqrt{35}\;[\;147+350+75\;]\\\)

Therefore, \(\left(\sqrt{7}+\sqrt{5} \right )^{6}-\left ( \sqrt{7}-\sqrt{5}\right)^{6}\)= \(2288\sqrt{35}\)

 

 

Q.6: Find the expansion of \(\left ( a^{3} +\sqrt{a^{3}-2}\right )^{4}+ \left ( a^{3} -\sqrt{a^{3}-2}\right )^{4}\)

 

Sol.

The above given expression can be assumed as: (x + y)4 + (x – y)4

Now, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ]  +  [ nC1 × (xn – 1) × y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  +  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

And, [x – y]n = [ nC0 × (xn) ]  –  [ nC1 × (xn – 1) ×  y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  –  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +  (-1)n [ nCn × yn ]

Therefore, [x + y]4 = [4C0 × (x4)]  +  [4C1 × (x3) × (y)]  +  [4C2 ×  (x2) ×  (y)2]  +  [4C3 × (x1) × (y)3] + [4C4 × (x0) × (y)4]

And, [x – y]4 = [4C0 × (x4) ]  –  [4C1 × (x3) × (y) ]  +  [4C2 ×  (x2) × (y)2 ]  –  [4C3 × (x1) × (y)3 ] + [4C4 × (x0) × (y)4 ]

Therefore, (x + y)4 + (x – y)4 = [ 4C0 x4 + 4C1 x3 y + 4C2 x2 y2 + 4C3 x y3 + 4C4 y4 ]  +  [ 4C0 x44C1 x3 y + 4C2 x2 y24C3 x y3 + 4C4 y4 ]

\(\\\Rightarrow\) [ 4C0 x4 + 4C1 x3y + 4C2 x2y2 + 4C3 xy3 + 4C4 y4 + 4C0 x44C1 x3y + 4C2 x2y24C3 xy3 + 4C4 y4 ]

\(\\\Rightarrow\)  2[ 4C0 x4 + 4C2 x2 y2 + 4C4 y4] = 2[ x4 + 6x2 y2 + y4 ]

Therefore, (x + y)4 + (x – y)4 = 2[ x4 + y4 + 6 x2 y2 ] . . . . . . . . . . . . . . . . . . . (1)

Now, on substituting x = a3 and y = \(\sqrt{a^{3}-2}\) in equation (1) we will get:

\(\\\Rightarrow \left ( a^{3} +\sqrt{a^{3}-2}\right )^{4}+ \left ( a^{3} -\sqrt{a^{3}-2}\right )^{4}:\\\)

\(\\=2\times \left [ (a^{3})^{4}+ (\sqrt{a^{3}-2})^{4}+[6\times (a^{3})^{2}\times (\sqrt{a^{3}-2})^{2}] \right ]\\\)

 \(\\=2\left [ a^{12}+ (a^{3}-2)^{2}+[6\times a^{6}\times (a^{3}-2)] \right ]\\\)

\(\\=2\left [\; a^{12}+ a^{6}+4-4a^{3}+6a^{9}-12a^{6} \;\right ]\\\)

\(\\\Rightarrow 2a^{12}+12a^{9}-22a^{6}-8a^{3}+8\)

Therefore, the expansion of \(\left ( a^{3} +\sqrt{a^{3}-2}\right )^{4}+ \left ( a^{3} -\sqrt{a^{3}-2}\right )^{4}\):

= \(2a^{12}+12a^{9}-22a^{6}-8a^{3}+8\)

 

 

Q.7: Find the approximate value of (0.98)5 using the first four terms of its expansion.

 

Sol.

(0.98)5 = (1 – 0.02)5

Now, by using Binomial Theorem:

Since, [ x – y ]n = [ nC0 × (x)n ]  –  [ nC1 × (x)n – 1 × y ]  +  [ nC2 ×  (x)n – 2 ×  y2 ]  –  [ nC3 × (x)n – 3 × y3 ] + . . . . . . . . . . . . . . . . . . (-1)n [ nCn × yn ]

Therefore, (1 – 0.02)5 = [ 5C0 × (1)5 ]  –  [ 5C1 × (1)4 × (0.02) ]  +  [ 5C2 ×  (1)3 ×  (0.02)2 ]  –  [ 5C3 × (1)2 × (0.02)3 ]

= 1 – [5 × 1 × 0.02] + [5 × 1 × 0.0004] – [5 × 1 × 0.000008]

= 1 – 0.10 + 0.0020 + 0.000040

= 0.90204

Therefore, (0.98)5 = 0.90204

 

 

Q.8: The coefficient of the 5th term from end and the 5th term from the beginning in the expansion of  \(\left ( 2^{\frac{1}{4}}+\frac{1}{3^{\frac{1}{4}}} \right )^{n}\) are in the ratio \(1:\sqrt{6}\).Find the value of ‘n’.

 

Sol.

Since, The general term in the expansion of (a + b)n  is given by: Tk + 1 = nCk × (a)n – k × bk

Now on substituting a = \(\left ( 2\right )^{{\frac{1}{4}}}\) and b = \(\frac{1}{(3)^{\frac{1}{4}}}\) in the above equation:

Tk + 1 = nCk ×\(\left ( \sqrt[4]{2} \right )^{n-k} \times \left ( \frac{1}{\sqrt[4]{3}}\right )^{k}\\\)

Now, 5th term from beginning in the expansion of \(\left ( 2^{\frac{1}{4}}+\frac{1}{3^{\frac{1}{4}}} \right )^{n}\):

T(4 + 1) = nC4 × \(\left ( \sqrt[4]{2} \right )^{n-4} \times \left ( \frac{1}{\sqrt[4]{3}} \right )^{4}\\\)

i.e.  T5 = nC4 × \(\left ( 2 \right )^{\frac{n}{4}-1} \times \left ( \frac{1}{3} \right )\\\)

\(\\\Rightarrow T_{5} = \frac{n!}{4!\;(n-4)!}\times (2)^{\frac{n}{4}}\times (2)^{-1}\times \frac{1}{3}\\\)

Therefore the coefficient of 5th term from beginning \(\boldsymbol{= \frac{n!}{4!\;(n-4)!}\times (2)^{\frac{n}{4}}\times \frac{1}{6}}\\\)

Now, 5th term from end in the expansion of \(\left ( 2^{\frac{1}{4}}+\frac{1}{3^{\frac{1}{4}}} \right )^{n}\):

1st term, 2nd term, 3rd term . . . . . . . . . . . (n – 5)th term, (n – 4)th term, (n – 3)th term, (n – 2)th term, (n – 1)th term, nth term.

Therefore, 5th term from end will be (n – 4)th term from beginning.

i.e.  T(n – 4) + 1 = nCn – 4 × \(\left ( \sqrt[4]{2} \right )^{n-(n-4)} \times \left ( \frac{1}{\sqrt[4]{3}} \right )^{n-4}\\\)

i.e.  T(n – 3) = nCn – 4 × \(\left ( \sqrt[4]{2} \right )^{4} \times \left ( \frac{1}{\sqrt[4]{3}} \right )^{n}\times \left ( \frac{1}{\sqrt[4]{3}} \right )^{-4}\\\)

\(\\\Rightarrow T_{(n-3)} = \frac{n!}{(n-4)!\;[(n-(n-4)]!}\times 2 \times \left ( \frac{1}{3} \right )^{\frac{n}{4}}\times 3\\\)

Therefore the coefficient of 5th term from end [ T(n – 4)th term ] \(= \frac{n!}{(n-4)!\;[(4)]!} \times \left ( \frac{1}{3} \right )^{\frac{n}{4}}\times 6\\\)

Now, according to the given conditions:

The ratio of coefficient of 5th term from end and the coefficient of 5th term from beginning is \(1:\sqrt{6}\\\)

Therefore, \(\\ \frac{T_{n-3}}{T_{5}}=\frac{1}{\sqrt{6}}\\\)

\(\\\Rightarrow \left ( \frac{n!}{(n-4)!\;[(4)]!} \times \left ( \frac{1}{3} \right )^{\frac{n}{4}}\times 6 \right )\div\left ( \frac{n!}{4!\;(n-4)!}\times (2)^{\frac{n}{4}}\times \frac{1}{6} \right )=\frac{1}{\sqrt{6}}\\\)

\(\\\Rightarrow \frac{4!\;(n-4)!}{(n-4)!\;4!}\times \frac{1}{(2)^{\frac{n}{4}}}\times \frac{1}{(3)^{\frac{n}{4}}}\times 6\times 6=\frac{1}{\sqrt{6}}\\\)

\(\\\Rightarrow \frac{1}{(6)^{\frac{n}{4}}}\times 36=\frac{1}{\sqrt{6}}\\\)

\(\\\Rightarrow 36\sqrt{6}=(6)^{\frac{n}{4}}\\\)

\(\\\Rightarrow (6)^{2}\times (6)^{\frac{1}{2}}=6^{\frac{n}{4}}\\\)

\(\\\Rightarrow (6)^{\frac{5}{2}}=(6)^{\frac{n}{4}\\}\)

On comparing RHS and LHS:

\(\\\Rightarrow \frac{5}{2}=\frac{n}{4}\\\)

\(\\\Rightarrow n = 10\)

Therefore, the value of n = 10.

 

 

Q.9: By using the Binomial Theorem, find the expansion of \(\left ( 1+\frac{x}{3}-\frac{3}{x} \right )^{4}\);     where 0

 

Sol.

The given expression \(\left ( 1+\frac{x}{3}-\frac{3}{x} \right )^{4}\) can be expanded as \(\left [\left (1-\frac{3}{x} \right )+\frac{x}{3} \right ]^{4}\)

Now, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ]  +  [ nC1 × (xn – 1) × y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  +  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

Therefore, \(\left [\left (1-\frac{3}{x} \right )+\frac{x}{3} \right ]^{4}\) = [4C0 × \(\left (1-\frac{3}{x} \right )^{4}\)]  +  [4C1 × \(\left (1-\frac{3}{x} \right )^{3}\) × \(\left ( \frac{x}{3} \right )\)]  +  [4C2 × \(\left (1-\frac{3}{x} \right )^{2}\) × \(\left ( \frac{x}{3} \right )^{2}\)]  +  [4C3 × \(\left (1-\frac{3}{x} \right )^{1}\) × \(\left ( \frac{x}{3} \right )^{3}\)  +  [4C4 × \(\left ( \frac{x}{3} \right )^{4}\)]

\(\\=\left [ 1\times \left (1-\frac{3}{x} \right )^{4} \right ]+\left [ 4 \times \left (1-\frac{3}{x} \right )^{3}\times \left ( \frac{x}{3} \right ) \right ]+\left [ 6\times \left (1-\frac{3}{x} \right )^{2} \times \left ( \frac{x}{3} \right )^{2} \right ]+\left [ 4 \times \left (1-\frac{3}{x} \right )^{1} \times \left ( \frac{x}{3} \right )^{3} \right ]+\left [ 1 \times \left ( \frac{x}{3} \right )^{4} \right ]\) . . . . . . . . . . . . . . . . . . . (1)

Now, \(\left (1-\frac{3}{x} \right )^{4}\):

By using Binomial Theorem:

Since, [x – y]n = [ nC0 × (xn) ]  –  [ nC1 × (xn – 1) × y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  –  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . (-1)n [ nCn × yn ]

Therefore, \(\left (1-\frac{3}{x} \right )^{4}\) = [4C0]  –  [4C1 × \(\left ( \frac{3}{x} \right )^{1}\)]  +  [4C2 × \(\left ( \frac{3}{x} \right )^{2}\)]  –  [4C3 × \(\left ( \frac{3}{x} \right )^{3}\)]  +  [4C4 × \(\left ( \frac{3}{x} \right )^{4}\)]

\(\\= \left [ 1 \right ] – \left [ 4 \times \left ( \frac{3}{x} \right ) \right ] + \left [ 6 \times \left ( \frac{9}{x^{2}} \right ) \right ] – \left [ 4\times \left ( \frac{27}{x^{3}} \right ) \right ] + \left [ 1 \times \left ( \frac{81}{x^{4}} \right ) \right ]\\\)

\(\\=\left [ 1-\frac{12}{x}+\frac{54}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}} \right ]\) . . . . . . . . (2)

And, \(\left (1-\frac{3}{x} \right )^{3}\) = [ 3C0 ]  –  [ 3C1 × \(\left ( \frac{3}{x} \right )^{1}\)]  +  [3C2 × \(\left ( \frac{3}{x} \right )^{2}\)]  –  [3C3 × \(\left ( \frac{3}{x} \right )^{3}\)]

\(\\= [1]-\left [ 3\times \left ( \frac{3}{x} \right ) \right ] + \left [ 3 \times \left ( \frac{9}{x^{2}} \right ) \right ] – \left [ 1\times \left ( \frac{27}{x^{3}} \right ) \right ]\\\)

\(\\=\left [1-\frac{9}{x}+\frac{27}{x^{2}}-\frac{27}{x^{3}} \right ]\) . . . . . . . . . . . . (3)

Now, from equation (1), equation (2) and equation (3):

\(\\\Rightarrow \left [ 1-\frac{12}{x}+\frac{54}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}}\right ]+\left [ 4 \times \left ( 1-\frac{9}{x}+\frac{27}{x^{2}}-\frac{27}{x^{3}} \right )\times \frac{x}{3} \right ]+\left [ 6\times \left (1+\frac{9}{x^{2}}-\frac{6}{x} \right )\times \frac{x^{2}}{9}\right ] +\left [ \left ( 4-\frac{12}{x} \right )\times \frac{x^{3}}{27} \right ]+\left [\frac{x^{4}}{81}\right ]\\\)

\(\\\Rightarrow \left [ 1-\frac{12}{x}+\frac{54}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}}\right ]+\left [ \frac{4x}{3}-12+\frac{36}{x}-\frac{36}{x^{2}}\right ]+\\\\\\\\+\left [ \frac{2x^{2}}{3}+6-4x\right ] +\left [ \frac{4x^{3}}{27}-\frac{4x^{2}}{9} \right ]+\left [\frac{x^{4}}{81}\right ]\\\)

\(\\\boldsymbol{\Rightarrow \left [ \frac{x^{4}}{81}+\frac{4x^{3}}{27}+\frac{2x^{2}}{9}-\frac{8x}{3}-5+\frac{24}{x}+\frac{18}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}}\right ]}\\\)

Therefore, the expansion of \(\left ( 1+\frac{x}{3}-\frac{3}{x} \right )^{4}\):

= \(\boldsymbol{\left [ \frac{x^{4}}{81}+\frac{4x^{3}}{27}+\frac{2x^{2}}{9}-\frac{8x}{3}-5+\frac{24}{x}+\frac{18}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}}\right ]}\)

 

 

Q.10: By using the Binomial Theorem, find the expansion of (6x2 – 4ax + 6a2)3.

 

Sol.

The given expression (6x2 – 4ax + 6a2)3 can be expanded as [(6x2 – 4ax) + 6a2)3]

By using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ]  +  [ nC1 × (xn – 1) × y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  +  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

Therefore, [(6x2 – 4ax) + 6a2)3] = [3C0 × (6x2 – 4ax)3]  +  [3C1 × (6x2 – 4ax)2 × 6a2]  +  [3C2 ×  (6x2 – 4ax)1 ×  (6a2)2]  +  [3C3 × (6x2 – 4ax)0 × (6a2)3]

= [1× (6x2 – 4ax)3]  +  [3 × (6x2 – 4ax)2 × 6a2]  +  [3 ×  (6x2 – 4ax) ×  (6a2)2]  +  [1 × (6a2)3]

= [(6x2 – 4ax)3]  +  [(6x2 – 4ax)2 × 18a2]  +  [(6x2 – 4ax) × 108a4 ]  +  [216a6] . . . . . . . . . . . . (1)

Now, [(6x2 – 4ax)3]:

By using Binomial Theorem:

Since, [x – y]n = [ nC0 × (xn) ]  –  [ nC1 × (xn – 1) × y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  –  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . (-1)n [ nCn × yn ]

Therefore, [(6x2 – 4ax)3] = [3C0 × (6x2)3]  –  [3C1 × (6x2)2) × 4ax]  +  [3C2 × 6x2 ×  (4ax)2]  –  [3C3 × (4ax)3]

= [1 × (216x6)]  –  [3 × (36x4) × 4ax]  +  [3 × (6x2 16a2x2]  –  [1 × (64a3x3)]

\(\Rightarrow\) [(6x2 – 4ax)3] = [216x6 – 432 ax5 + 288 a2x4 – 64a3x3]  . . . . . . . . . . (2)

From equation (1) and equation (2)

=[216x6 – 432 ax5 + 288 a2x4 – 64a3x3] + [(36x4 + 16a2x2 – 48ax3) × 18a2] + [(6x2 – 4ax) × 108a4 ] + [ 216a6 ]

=[216x6 – 432 ax5 + 288 a2x4 – 64a3x3 + 648a2x4 + 288a4x2 – 864a3x3 + 648x2a4 – 432xa5 + 216a6]

=8[27x6 – 54 ax5 + 36 a2x4 – 8a3x3 + 81a2x4 + 36a4x2 – 108a3x3 + 81x2a4 – 54xa5 + 27a6]

=8[27x6 – 54 ax5 + 117a2 x4 – 116a3x3 + 117a4x2 – 54a5x + 27a6]

Therefore, the expansion of (6x2 – 4ax + 6a2)3 :

= 8[27x6 – 54 ax5 + 117a2 x4 – 116a3x3 + 117a4x2 – 54a5x + 27a6]

 

 

Q.11: By using the Binomial Theorem, find the expansion of \(\left ( 1+\frac{x}{5}-\frac{5}{x} \right )^{4}\); where 0

 

Sol.

The given expression \(\left ( 1+\frac{x}{5}-\frac{5}{x} \right )^{4}\) can be expanded as \(\left [\left (1-\frac{5}{x} \right )+\frac{x}{5} \right ]^{4}\)

Now, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ]  +  [ nC1 × (xn – 1) × y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  +  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

Therefore, \(\left [\left (1-\frac{5}{x} \right )+\frac{x}{5} \right ]^{4}\)

= [4C0 × \(\left (1-\frac{5}{x} \right )^{4}\)]  +  [4C1 × \(\left (1-\frac{5}{x} \right )^{3}\) × \(\left ( \frac{x}{5} \right )\)]  +  [4C2 × \(\left (1-\frac{5}{x} \right )^{2}\) × \(\left ( \frac{x}{5} \right )^{2}\)]  +  [4C3 × \(\left (1-\frac{5}{x} \right )^{1}\) × \(\left ( \frac{x}{5} \right )^{3}\)  +  [4C4 × \(\left ( \frac{x}{5} \right )^{4}\)]

\(\\=\left [ 1\times \left (1-\frac{5}{x} \right )^{4} \right ]+\left [ 4 \times \left (1-\frac{5}{x} \right )^{3}\times \left ( \frac{x}{5} \right ) \right ]+\left [ 6\times \left (1-\frac{5}{x} \right )^{2} \times \left ( \frac{x}{5} \right )^{2} \right ]+\left [ 4 \times \left (1-\frac{5}{x} \right )^{1} \times \left ( \frac{x}{5} \right )^{3} \right ]+\left [ 1 \times \left ( \frac{x}{5} \right )^{4} \right ]\) . . . . . . . . . . . . . . . . . . . (1)

Now, \(\left (1-\frac{5}{x} \right )^{4}:\\\)

By using Binomial Theorem:

Since, [x – y]n = [ nC0 × (xn) ]  –  [ nC1 × (xn – 1) × y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  –  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . (-1)n [ nCn × yn ]

Therefore, \(\left (1-\frac{5}{x} \right )^{4}\) = [4C0]  –  [4C1 × \(\left ( \frac{5}{x} \right )^{1}\)]  +  [4C2 × \(\left ( \frac{5}{x} \right )^{2}\)]  –  [4C3 × \(\left ( \frac{5}{x} \right )^{3}\)]  +  [4C4 × \(\left ( \frac{5}{x} \right )^{4}\)]

\(\\= \left [ 1 \right ] – \left [ 4 \times \left ( \frac{5}{x} \right ) \right ] + \left [ 6 \times \left ( \frac{25}{x^{2}} \right ) \right ] – \left [ 4\times \left ( \frac{125}{x^{3}} \right ) \right ] + \left [ 1 \times \left ( \frac{625}{x^{4}} \right ) \right ]\\\)

\(\\=\left [ 1-\frac{20}{x}+\frac{150}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}} \right ]\) . . . . . . . . (2)

And, \(\left (1-\frac{5}{x} \right )^{3}:\\\)

[ 3C0 ]  –  [ 3C1 × \(\left ( \frac{5}{x} \right )^{1}\)]  +  [3C2 × \(\left ( \frac{5}{x} \right )^{2}\)]  –  [3C3 × \(\left ( \frac{5}{x} \right )^{3}\)]

\(\\= [1]-\left [ 3\times \left ( \frac{5}{x} \right ) \right ] + \left [ 3 \times \left ( \frac{25}{x^{2}} \right ) \right ] – \left [ 1\times \left ( \frac{125}{x^{3}} \right ) \right ]\\\)

\(\\=\left [ 1-\frac{15}{x}+\frac{75}{x^{2}}-\frac{125}{x^{3}} \right ]\) . . . . . . . . . . . . . . . . . . . (3)

Now, from equation (1), equation (2) and equation (3):

\(\\\Rightarrow \left [ 1-\frac{20}{x}+\frac{150}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}} \right ]+\left [ 4 \times \left [ 1-\frac{15}{x}+\frac{75}{x^{2}}-\frac{125}{x^{3}} \right ] \times \left ( \frac{x}{5} \right ) \right ]+\\\\\\+\left [ 6\times \left (1+\frac{25}{x^{2}} -\frac{10}{x}\right ) \times \left ( \frac{x}{5} \right )^{2} \right ]+\left [ 4 \times \left (1-\frac{5}{x} \right )\times \left ( \frac{x}{5} \right )^{3} \right ]+\left [ \left ( \frac{x}{5} \right )^{4} \right ]\\\)

\(\\\Rightarrow \left [ 1-\frac{20}{x}+\frac{150}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}} \right ]+\left [ \frac{4x}{5}-12+\frac{60}{x}-\frac{100}{x^{2}} \right ]+\left [ \frac{6x^{2}}{25}+6-\frac{12x}{5}\right ]+\left [ \frac{4x^{3}}{125}-\frac{x^{2}}{25} \right ]+\left [ \frac{x^{4}}{625} \right ]\\\)

\(\\\\\Rightarrow \left [\frac{50}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}}-\frac{8x}{5}-5+\frac{40}{x}+ \frac{x^{2}}{5}+\frac{4x^{3}}{125}+\frac{x^{4}}{625} \right ]\\\\\)

Therefore, the expansion of \(\left ( 1+\frac{x}{5}-\frac{5}{x} \right )^{4}\):

= \(\left [\frac{x^{4}}{625}+\frac{4x^{3}}{125}+ \frac{x^{2}}{5}-\frac{8x}{5}-5+\frac{40}{x}+\frac{50}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}}\right ]\)

 

 

Q.12: By using the Binomial Theorem, find the expansion of (5x3 – 2ax + 3a3)3.

 

Sol.

The given expression (5x3 – 2ax + 3a3)3 can be expanded as [(5x3 – 2ax) + 3a3)3]

By using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ]  +  [ nC1 × (xn – 1) × y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  +  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

Therefore, [ (5x3 – 2ax) + 3a3)3 ] = [ 3C0 × (5x3 – 2ax)3 ]  +  [ 3C1 × (5x3 – 2ax)2 × 3a3 ]  +  [ 3C2 ×  (5x3 – 2ax)1 ×  (3a3)2 ]  +  [ 3C3 × (5x3 – 2ax)0 × (3a3)3 ]

= [1× (5x3 – 2ax)3]  +  [3 × (5x3 – 2ax)2 × 3a3]  +  [3 ×  (5x3 – 2ax) ×  (3a3)2]  +  [1 × (3a3)3]

= [(5x3 – 2ax)3]  +  [(5x3 – 2ax)2 × 9a3]  +  [(5x3 – 2ax) × 27a6]  +  [27a9] . . . . . . . . . . . . (1)

Now, [(5x3 – 2ax)3]:

By using Binomial Theorem:

Since, [x – y]n = [ nC0 × (xn) ]  –  [ nC1 × (xn – 1) × y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  –  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . (-1)n [ nCn × yn ]

Therefore, [(5x2 – 2ax)3] = [3C0 × (5x3)3]  –  [3C1 × (5x3)2) × 2ax]  +  [3C2 × 5x3 ×  (2ax)2]  –  [3C3 × (2ax)3]

= [1 × 125x9]  –  [3 × (25x6)× 2ax]  +  [3 × (5x3) ×  4a2x2]  –  [1 × 8a3x3]

[(5x3 – 2ax)3] = [125x6 – 150ax5 + 60a2x4 – 8a3x3] . . . . . . . . . . (2)

From equation (1) and equation (2):

=[125x6 – 150ax5 + 60a2x4 – 8a3x3] + [(25x6 + 4a2x2 – 20ax4) × 9a3]  + [(5x3 – 2ax) × 27a6 ] + [ 27a9 ]

=[125x6 – 150ax5 + 60a2x4 – 8a3x3] + [225a3x6 + 36a5x2 – 180a4x4] + [135a6x3 – 54a7x ] + [ 27a9 ]

=[125x6 – 150ax5 + 60a2x4 – 8a3x3 + 225a3x6 + 36a5x2 – 180a4x4 + 135a6x3 – 54a7x  +  27a9 ]

=[125x6 + 225a3x6 – 150ax5 + 60a2x4 – 180a4x4 – 8a3x3 + 135a6x3 + 36a5x2 – 54a7x  +  27a9 ]

 Therefore, the expansion of (5x3 – 2ax + 3a3)3 :

=[125x6 + 225a3x6 – 150ax5 + 60a2x4 – 180a4x4 – 8a3x3 + 135a6x3 + 36a5x2 – 54a7x + 27a9]

 

 

Q.13: Find the coefficient of x7 in the product of (1 + x)7 (1 + 3x)6 by using binomial theorem.

 

Sol.

By using Binomial Theorem:

Since, (1 + x)n = [ nC0 ]  +  [ nC1 × (x) ]  +  [ nC2 × (x)2 ]  +  [ nC3 × (x3) ] + . . . . . . . . . . . . . . . . . . + [ nCn × (x)n ]

Therefore, (1 + x)7 = [ 7C0 ]  +  [ 7C1 × (x)  ]  +  [ 7C2 × (x)]  +  [ 7C3 × (x)]  + [ 7C4 × (x)4 ]  +  [ 7C5 × (x5) ]  +  [ 7C6 × (x)6 ]  +  [ 7C7 × (x)7

\(\\\Rightarrow\) 1 + [ 7x ] + [ 21x2 ] + [ 35x3 ] + [ 35x4 ] + [ 21x5 ] + [ 7x6 ] + [ x7 ]

Therefore, (1 + x)7 = 1+7x + 21x2 + 35x3 + 35x4 + 21x5 + 7x6 + x7

Now, (1 + 3x)6:

By using Binomial Theorem:

(1 + 3x)6 = [ 6C0 ]  +  [ 6C1 × (3x)  ]  +  [ 6C2 × (3x)]  +  [ 6C3 × (3x)]  + [ 6C4 × (3x)4 ]  +  [ 6C5 × (3x5) ]  +  [ 6C6 × (3x)6 ]

\(\\\Rightarrow\) 1 + [ 6 × (3x) ] + [ 15 × (9x2) ] + [ 20 × (27x3) ] + [ 15 × (81x4) ] + [ 6 × (243x5) ] + [ 1 × 729x6 ]

\(\\\Rightarrow\) 1 + 18x + 135x2 + 540x3 + 1215x4 + 1458x5 + 729x6

Therefore, (1 + 3x)6 = 1 + 18x + 135x2 + 540x3 + 1215x4 + 1458x5 + 729x6

Now, (1 + x)7 (1 + 3x)6:

(1+7x + 21x2 + 35x3 + 35x4 + 21x5 + 7x6 + x7) × (1 + 18x + 135x2 + 540x3 + 1215x4 + 1458x5 + 729x6)

Now, it not required to multiply the above equation, only terms with x5 are required and that can be identified by analysing the above equation.

\(\\\Rightarrow\) [{(7x) × (729x6)} + {(21x2) × (1458x5)} + {(35x3) × (1215x4)} + {(35x4) × (540x3)} + {(21x5) × (135x2)} + {(7x6) × (18x)} + {(x7) × (1)}]

\(\\\Rightarrow\) [5103x5 + 30618x5 + 42525x5+ 18900x5 + 2835x7 + 126x7 + x7]

\(\\\Rightarrow\\\) 72551 x7

Therefore, the coefficient of x7 = 72551

 

 

Q.14:  Evaluate \(\left(\sqrt{3}+\sqrt{7} \right )^{5}-\left ( \sqrt{3}-\sqrt{7}\right)^{5}\)

 

Sol.

Find (a + b)5 – (a – b)5

Now, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ]  +  [ nC1 × (xn – 1) × y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  +  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

And, [x – y]n = [ nC0 × (xn) ]  –  [ nC1 × (xn – 1) ×  y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  –  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +  (-1)n [ nCn × yn ]

Therefore, [a + b]5 = [ 5C0 × (a5) ] + [ 5C1 × (a4) × b ] + [ 5C2 ×  (a3) ×  b2 ] + [ 5C3 × (a2) × b3 ] + [ 5C4 × (a1) × b4 ] + [ 5C5 × b5 ]

And, [a – b]5 = [ 5C0 × (a5) ] – [ 5C1 × (a4) × b ] + [ 5C2 ×  (a3) ×  b2 ] – [ 5C3 × (a2) × b3 ] + [ 5C4 × (a) × b4 ] – [ 5C5 × b5 ]

Therefore, (a + b)5 – (a – b)5 = [5C0 a5 + 5C1 a4b + 5C2 a3b2 + 5C3 a2b3 + 5C4 ab4 + 5C5 b5] – [5C0 a55C1 a4 b + 5C2 a3b2 5C3 a2b3 + 5C4 ab4 5C5 ab5]

\(\\\Rightarrow\)  [5C0 a5 + 5C1 a4b + 5C2 a3b2 + 5C3 a2b3 + 5C4 ab4 + 5C5 b55C0 a5 + 5C1 a4 b – 5C2 a3b2 + 5C3 a2b3 5C4 ab4 + 5C5 ab5]

\(\\\Rightarrow\)  2[ 5C1 a4b + 5C3 a2b3 + 5C5 b5 ] = 2[ 5a4b + 10a2b3 + b5 ]

Therefore, (a + b)5 – (a – b)5 = 2b[ 5a4 + 10a2 b2 + b4 ] . . . . . . (1)

Now, on substituting a = \(\sqrt{3}\) and b = \(\sqrt{7}\) in equation (1) we will get:

\(\\\left(\sqrt{3}+\sqrt{7} \right )^{5}-\left ( \sqrt{3}-\sqrt{7}\right)^{5}:\\\)

\(\\=2(\sqrt{7})\times \left [ 5( \sqrt{3})^{4} +10\times ( \sqrt{3})^{2}\times (\sqrt{7})^{2}+(\sqrt{7})^{4} \right ]\\\)

\(\\\Rightarrow 2\sqrt{7}\left [\;45+210+49\; \right ]\\\)

Therefore, \(\left(\sqrt{3}+\sqrt{7} \right )^{5}-\left ( \sqrt{3}-\sqrt{7}\right)^{5}\)= \(608\sqrt{7}\)

 

 

Q.15: Find the expansion of \(\left ( a^{2} +\sqrt{a^{2}-5}\right )^{4}+ \left ( a^{2} -\sqrt{a^{2}-5}\right )^{4}\)

 

Sol.

The above given expression can be assumed as: (x + y)4 + (x – y)4

Now, by using Binomial Theorem:

Since, [x + y]n = [ nC0 × (xn) ]  +  [ nC1 × (xn – 1) × y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  +  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]

And, [x – y]n = [ nC0 × (xn) ]  –  [ nC1 × (xn – 1) ×  y ]  +  [ nC2 ×  (xn – 2) ×  y2 ]  –  [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +  (-1)n [ nCn × yn ]

Therefore, [x + y]4 = [4C0 × (x4)]  +  [4C1 × (x3) × (y)]  +  [4C2 ×  (x2) ×  (y)2]  +  [4C3 × (x1) × (y)3] + [4C4 × (x0) × (y)4]

And, [x – y]4 = [4C0 × (x4) ]  –  [4C1 × (x3) × (y) ]  +  [4C2 ×  (x2) × (y)2 ]  –  [4C3 × (x1) × (y)3 ] + [4C4 × (x0) × (y)4 ]

Therefore, (x + y)4 + (x – y)4 = [ 4C0 x4 + 4C1 x3 y + 4C2 x2 y2 + 4C3 x y3 + 4C4 y4 ]  +  [ 4C0 x44C1 x3 y + 4C2 x2 y24C3 x y3 + 4C4 y4 ]

\(\\\Rightarrow\) [ 4C0 x4 + 4C1 x3y + 4C2 x2y2 + 4C3 xy3 + 4C4 y4 + 4C0 x44C1 x3y + 4C2 x2y24C3 xy3 + 4C4 y4 ]

\(\\\Rightarrow\)  2[ 4C0 x4 + 4C2 x2 y2 + 4C4 y4] = 2[ x4 + 6x2 y2 + y4 ]

Therefore, (x + y)4 + (x – y)4 = 2[ x4 + y4 + 6 x2 y2 ] . . . . . . . . . . . . . . . . . . . (1)

Now, on substituting x = a2 and y = \(\sqrt{a^{2}-5}\) in equation (1) we will get:

\(\\\Rightarrow \left ( a^{2} +\sqrt{a^{2}-5}\right )^{4}+ \left ( a^{2} -\sqrt{a^{2}-5}\right )^{4}:\\\)

\(\\=2\times \left [ (a^{2})^{4}+ \left ( \sqrt{a^{2}-5} \right )^{4}+6\times (a^{2})^{2}\times \left ( \sqrt{a^{2}-5} \right )^{2} \right ]\\\)

\(\\\Rightarrow 2\left [ a^{8}+ (a^{2}-5)^{2}+[6a^{4}\times (a^{2}-5)] \right ]\\\)

\(\\\Rightarrow 2 \left [ a^{8}+a^{4}+ 25 + 10a^{2}+6a^{6}-30a^{4} \right ]\\\)

Therefore, the expansion of \(\left [ a^{2} +\sqrt{a^{2}-5}\right ]^{4}+ \left [ a^{2} -\sqrt{a^{2}-5}\right ]^{4}:\)

=\(2 \left [ a^{8}+6a^{6}-29a^{4}+10a^{2}+25 \right ]\)

These were some of the important uses of Binomial theorem. For the students of class 11, it is essential to have the basic conceptual knowledge of the Binomial Theorem. Students can get to learn the various concepts of Algebra and even derive various formulas with the help of Binomial Theorem. It is essential for the students to have a better understanding of the topic so as to excel in the examination.

Below given are the NCERT Solutions Class 11 Maths Binomial Theorem, which needs to be practiced before the examination.

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