NCERT Solutions For Class 11 Maths Chapter 8

NCERT Solutions Class 11 Maths Binomial Theorem

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 NCERT Solutions 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 ]

 

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 = nr=0 nCr (a)n – r × br

 

NOTE:

 

(1.) nCr = n!r!(nr)! 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 ]

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

1 – 15x + 90x2 – 270x3 + 405x4 – 243x5

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

 

 

Q.2: Expand the Expression (5xx5)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, (5xx5)5:

[5C0 × (5x)5 ] – [5C1 × (5x)4 ×  (x5)1 ] + [5C2 ×  (5x)3 × (x5)2 ] – [ 5C3 × (5x)2 × (x5)3 ] + [5C4 × (5x)1 × (x5)4 ] – [5C5 × (x5)5]

[1×3125x5][5×625x4×x5]+[10×125x3×x225][10×25x2×x3125]+[5×5x×x4625][1×x53125] [3125x5][625x3]+[50x]2x+[3x325][x53125]

Therefore, (5xx5)5:

[x53125]+[3x325]2x+[50x][625x3]+[3125x5]

 

 

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 ]

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

729x6 – 2916x5 + 4860x4 – 4320x3 + 2160x2 – 576x + 64

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

 

 

Q.4: Expand the Expression (4x+x3)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, (4x+x3)5:

[ 5C0 × (4x)5 ]  +  [ 5C1 × (4x)4 ×  (x3)1 ]  +  [ 5C2 ×  (4x)3 × (x3)2 ]  +  [ 5C3 × (4x)2 × (x3)3 ]  +  [ 5C4 × (4x)1 × (x3)4 ]  +  [ 5C5 × (x3)5]

[1×1024x5]+[5×256x4×x3]+[10×64x3×x29]+[10×16x2×x327]+[5×4x×x481]+[1×x5243] [1024x5]+[12803x3]+[6409x]+[160x27]+[20x381]+[x5243]

Therefore, (4x+x3)5:

[x5243]+[20x381]+[160x27]+[6409x]+[12803x3]+[1024x5]

 

 

Q.5: Expand the Expression (1xx2)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, (1xx2)6:

[ 6C0 × (1x)6 ]  –  [ 6C1 × (1x)5 ×  (x2)1 ]  +  [ 6C2 ×  (1x)4 × (x2)2 ]  –  [ 6C3 × (1x)3 × (x2)3 ]  +  [ 6C4 × (1x)2 × (x2)4 ]  –  [ 6C5 × (1x)1 × (x2)5 ]  +  [ 6C6 ×(x2)6]

[1×1x6][6×1x5×x2]+[15×1x4×x24][20×1x3×x38]+[15×1x2×x416][6×1x×x532]+[1×x664] [1x6][3x4]+[154x2][52]+[15x216][3x416]+[x664]

Therefore, (1xx2)6:

[x664][3x416]+[15x216][52]+[154x2][3x4]+[1x6]

 

 

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]

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

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]

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

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 ]

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

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 ]

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

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

   [1 × 1] + [10000 × 1 × 0.1] + other positive terms

   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 (5+3)5(53)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 ]

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

 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 = 5 and b = 3 in equation (1) we will get:

(5+3)5(53)5=23×[5(5)4+10(5)2×(3)2+(3)4]

23[125+150+9]=5683

Therefore, (5+3)5(53)5=5683

 

 

Q.12:  Find (1 + x)5 – (1 – x)5 . Hence evaluate (1+7)5(17)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 ]

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

  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 = 7 in equation (1) we will get:

(1+7)7(17)5=27[5+10(7)2+(7)4] 27[5+70+49]=2487

Therefore, (1+7)5(17)5=2487

 

 

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 ]

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

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

   (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 nr=0 2r × nCr = 3n

 

Sol.

By using Binomial Theorem:

Since, (a + b)n = nr=0 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 = nr=0 nCr (1)n – r × (2)r

3n = nr=0 nCr × (2)r

Therefore, nr=0 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 ]

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

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

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: (2+3)6(23)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 ]

 [ 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 ]

  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 = 2 and b = 3 in equation (1) we will get:

(2+3)6(23)6=4(2.3)[3(2)4+10(2)2×(3)2+3(3)4] 46[12+60+27]=3966

Therefore, (2+3)6(23)6=3966

 

 

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]

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

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]

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

10000000000 + 3500000000 + 490000000 + 34300000 + 1200500 + 16807

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

 

 

Q.19: Expand the Expression (32xx5)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, (32xx5)6:

[6C0 × (32x)6 ]  –  [6C1 × (32x)5 ×  (x5)1 ]  +  [6C2 ×  (32x)4 × (x5)2 ]  –  [6C3 × (32x)3 × (x5)3 ]  +  [6C4 × (32x)2 × (x5)4 ]  –  [6C5 × (32x)1 × (x5)5 ]  +  [6C6 ×(x5)6]

[1×72964×x6][6×24332×x5×x5]+[15×8116×x4×x225][20×278x3×x3125]+[15×94x2×x4625][6×32x×x53125]+[1×x615625] [72964x6][72980x4]+[24380x2][2750]+[27x2500][9x43125]+[x615625]

Therefore, (32xx5)6:

[x615625][9x43125]+[27x2500][2750]+[24380x2][72980x4]+[72964x6]

 

 

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 ]

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

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 = (n2+1)thterm

 

    (b).   If n is odd:

The middle term = (n+12)thtermand(n+12+1)thterm    

 

 

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

T4=9!3!6!×x6×8 T4=9×8×7×6!3×2×1×6!×8x6=672x6

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

T9=14!8!6!×64×729×a6b8 T9=14×13×12×11×10×9×8!6×5×4×3×2×1×8!×64×729×a6b8 T9=3003×64×729×a6b8=140107968a6b8

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

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

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

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:

   T3 + 1 = 10C3 × (2x)10 – 3 × (3y)3

   T4 = 10C3 × (2x)7 × (3y)3

T4=10!3!7!×128x7×27y3 T4=10×9×8×7!3×2×1×7!×3456x7y3 T4=120×3456x7y3=414720x7y3

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

 

 

Q.6 Find the 11th term in the expansion of (8x12x)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 = (12x) we will get:

   T10 + 1 = 15C10 × (8x)15 – 10 × (12x)10

   T11 = 15C10 × (8x)5 × (12x)10

T11=15!10!5!×8×8×8×8×8×x5×1210×(1x)10 T11=15×14×13×12×11×10!5×4×3×2×1×10!×32×x5×1x5 T11=3003×32=96096

Therefore, 11th term in the expansion of (8x12x)15= 96096

 

 

Q.7: In the expansion of (5x410)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 = (n+12)thtermand(n+12+1)thterm

Therefore, the middle terms in the expansion of (5x410)9are:

(9+12)thtermand(9+12+1)thterm

5th term and 6th term

Now, 5th and 6th terms in the expansion of (5x410)9 are:

T5 = T4 + 1 = 9C4 × (5)9 – 4 × (x410)4

   T5 = 9C4 × (5)5 × (x410)4

T5=9!4!5!×(5)5×(x410)4 T5=9×8×7×6×5!4×3×2×1×5!×5×5×5×5×5×1104×x16 T5=126×516x16=3158x16

Therefore, 5th term =3158x16

Now, T6 = T5 + 1 = 9C5 × (5)9 – 5 × (x410)5

   T6 = 9C5 × (5)4 × (x410)4

T6=9!5!4!×(5)4×(x410)5 T6=9×8×7×6×5!5!×(4×3×2×1)×5×5×5×5×1105×x20 T6=126×1320x20=63160x20

Therefore, 6th term =63160x20

Therefore, 5th and 6th terms are the middle terms in the expansion of (5x410)9

And also, 5th term =3158x16 and 6th term =63160x20

 

 

Q.8: In the expansion of (x2+8y)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:

(n2+1)thterm

Therefore, the middle term in the expansion of (x2+8y)10 is:

(102+1)thterm = 6th term

Now, 6th term in the expansion of (x2+8y)10 is:

T6 = T5 + 1 = 10C5 × (x2)105 × (8y)5

   T6 = 10C5 × (x2)5 × (8y)5

T6=10!5!5!×(x2)5×(8y)5 T6=10×9×8×7×6×5!5×4×3×2×1×5!×x532×8×8×8×8×8×y5 T6=