09243500460  09243500460

Chapter 8: Binomial Theorem

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

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



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