Binomial Theorem – Exercise 8.1 – Class XI

  1. Expand the expression (1 – 2x)5

Solution:

By using Binomial theorem the expression (1-2x)5 can be expected as

(1 – 2x)5

= 5C0(1)55C1(1)4 (2x)15C2(1)3(2x)2 5C3(1)2(2x)3 5C4(1)1(2x)45C5(2x)5

= 1 – 5(2x) + 10(4x2) – 10(8x3) + 5(16x4) – (32x5)

= 1 – 10x + 40x2 – 80x3 + 80x4 – 32x5


  1. Expand the expression (2/xx/2)5

Solution:

By using Binomial theorem the expression (2/xx/2)5can be expected as

(2/xx/2)5

= 5C0(2/x)55C1(2/x)4 (x/2)15C2(2/x)3(x/2)2 5C3(2/x)2(x/2)3 5C4(2/x)1(x/2)45C5(x/2)5

= 32/x5 – 5(16/x4)(x/2) + 10(8/x3)(x2/4) – 10(4/x2)(x3/8)+5(2/x)(x4/16) – (x5/32)

= 32/x540/x3 + 20/x – 5x + 5/8 x3 – (x3/32)


  1. Expand the expression (2x – 3)6

Solution:

By using Binomial theorem the expression (2x – 3)6 can be expected as

(2x – 3)6

= 5C0(2x)55C1(2x)4 (3)15C2(2x)3(3)2 5C3(2x)2(3)3 5C4(2x)1(3)45C5(3)5

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

= 64x5 – 57x5 + 2160x4 – 4320x3 + 4860x2 – 2916x + 729


  1. Expand the expression (x/3 + 1/x)5

Solution:

By using Binomial theorem the expression (x/3 + 1/x)5

can be expected as

(x/3 + 1/x)5

= 5C0(x/3)55C1(x/3)4 (1/x)15C2(x/3)3(1/x)2 5C3(x/3)2(1/x)3 5C4(x/3)1(1/x)45C5(1/x)5

= (x5/243)  +  5(x4/81)(1/x) + 10(x3/27)(1/x2) + 5(x2/9)(1/x3)+5(x/3)(1/x4) + (1/x5)

= (x5/243) + (5x3/81) + (10x/27) + (10/9x) + (5/3x3) + (1/x5)


  1. Expand the expression (x + 1/x)6

Solution:

By using Binomial theorem the expression (x + 1/x)6

can be expected as

(x + 1/x)6

= 5C0x55C1 x4 (1/x)15C2 x3(1/x)2 5C3x2(1/x)3 5C4x1(1/x)45C5(1/x)5

= x6 +6(x)5(1/x) + 15(x)4 (1/x2) + 20(x)3(1/x3) + 15(x)2(1/x4)+6(x)(1/x5) + (1/x6)

= x6 + 6x4 + 15x2 + 20 + (15/x2) + (6/x4) + (1/x6)


  1. Using binomial theorem, evaluate (96)3

Solution:

96 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then binomial theorem can be applied.

(96)3 = (100 – 4)3

= 3C0 (100)33C1 (100)2 4 + 3C2 (100)(4)23C3 (4)3

= (100)3 – 3(100)2 4 +3(100)(4)2 – (4)3

= 1000000 – 120000 + 4800 – 64

= 884736


  1. Using binomial theorem, evaluate (102)5

Solution:

102 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then binomial theorem can be applied.

It can be written that, 102 = 100 + 2

(102)5 = (100 + 2)5

= 5C0 (100)5 + 5C1 (100)4 2 + 5C2 (100)3(2)2 + 5C3 (100)2(2)3 + 5C4 (100)1(2)4 + + 5C5(2)5

= (100)5 + 5(100)4 2 +10 (100)3(2)2 + 10(100)2(2)3 + 5(100)(2)4 + (2)5

= 11040808032


  1. Using binomial theorem, evaluate (101)4

Solution:

101 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then binomial theorem can be applied.

It can be written that, 101 = 100 + 1

(101)4 = (100 + 1)4

= 4C0 (100)4 + 4C1 (100)3 (1) + 4C2 (100)2(1)2 + 4C3 (100)(1)3 + 4C4 (1)4

= (100)4 +4(100)3 + 6(100)2 +4(100)+(1)4

=104060401


  1. Using binomial theorem, evaluate (99)5

Solution:

99 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then binomial theorem can be applied.

It can be written that, 99 = 100 – 1

(99)5 = (100 – 1)5

= 5C0 (100)55C1 (100)4 (1) + 5C2 (100)3(1)25C3 (100)2(1)3 + 5C4 (100)(1)45C5(1)5

= (100)5 – 5(100)4 + 10(100)3 – 10(100)2+ 5(100) – 1

=9509900499


  1. Using binomial theorem , indicate which number is larger (1.1)10000 or 1000

Solution:

By spitting 1.1 and then applying binomial theorem, the first few terms of (1.1)10000 can be obtained as

(1.1)10000  = (1 + 0.1)10000

= 10000C0 + 10000C1(1.1) + (other positive temrs)

= 1 + 10000 x 1.1 + (other positive temrs)

>1000

Hence, (1.1)10000 > 1000


  1. Find (a + b)4 – (a – b)4 Hence, evaluate (√3+ √2)4– (√3 – √2)4

Solution:

Using binomial theorem the expressions (a + b)4 and (a – b)4 can be expressed as

(a + b)4 = 4C0 a4 + 4C1 a3b + 4C2 a2b2 + 4C3 ab3 +4C4 b4

(a – b)4 = 4C0 a44C1 a3b + 4C2 a2b24C3 ab3 +4C4 b4

(a + b)4 – (a – b)4 = [4C0 a4 + 4C1 a3b + 4C2 a2b2 + 4C3 ab3 +4C4 b4] – [4C0 a44C1 a3b + 4C2 a2b24C3 ab3 +4C4 b4]

= 2(4C1 a3b + 4C3ab3)

= 2(4a2b + 4ab3)

= 8ab(a2 + b2)

By putting a = √3 and b = √2, we obtain

(√3+√2)4 – (√3-√2)4 = 8(√3)( √2){(√3)2 + (√2)2}

= 8(√6){3 + 2}

= 40√6


  1. Find (x+1)6 + (x – 1)6. Hence or otherwise evaluate (√2 + 1)6 + (√2 – 1)6

Solution:

Using Binomial theorem the expression (√2 + 1)6 + (√2 – 1)6 can be expressed as

(x + 1)6 = 6C0 x6 + 6C1 x5 + 6C2 x412 + 6C3 x3(1)3 +6C4 x2(1)4 + 6C5x(1)5 + 6C6 (1)6

(x – 1)6 = 6C0 x66C1 x5 + 6C2 x4126C3 x3(1)3 +6C4 x2(1)46C5x(1)5 + 6C6 (1)6

Therefore, (x + 1)6 – (x – 1)6 ={6C0 x6 + 6C1 x5 + 6C2 x412 + 6C3 x3(1)3 +6C4 x2(1)4 + 6C5x(1)5 + 6C6 (1)6} – {6C0 x66C1 x5 + 6C2 x4126C3 x3(1)3 +6C4 x2(1)46C5x(1)5 + 6C6 (1)6}

= 2 {6C0x6 + 6C2x4 +6C4x2 +6C6 }

= 2{x6 + 15x4 +15x2 + 1}

By putting x = (√2), we obtain

(√2 + 1)6 + (√2 – 1)6 = 2{(√2)6 + 15(√2)4 +15(√2)2 + 1}

= 2(8 + 15×4 + 15×2 + 1)

= 2(8 + 60 + 30 + 1)

= 2(99)

=198


  1. Show that 9n+1 – 8n – 9 is divisible by 64, whenever n is a positive integer.

Solution:

In order to show that 9n+1 – 8n – 9 is divisible by 64, it has to be proved that 9n+1 – 8n – 9 = 64k, where k is some natural number

By binomial theorem

(1 + a)m = mC0  + mC1 a + mC2 a2 + …+mCm am

For a = 8 and m = n + 1, we obtain

(1 + 8)n+1 = n+1C0 + n+1C1(8)+ n+1C2(8)2+…+ n+1Cn+1(8)n+1

9n+1 = 1 + (n+1)(8)+82[n+1C2 + n+1C3 8+ …+ n+1Cn+18n-1]

9n+1=9+8n+64[n+1C2 + n+1C3 x 8 + …+n+1Cn+1(8)n-1]

9n+1 – 8n – 9 = 64k, where k = n+1C2 + n+1C3x8 + …+n+1Cn+1(8)n-1 is a natural number

Thus 9n+1 – 8n – 9 is divisible by 64, whenever n is a positive integer.


  1. Prove that

Binomial Theorem – Exercise 8.1 – Class XI