Class 11 Maths NCERT Solutions for Chapter 8 Binomial Theorem Exercise 8.1
Binomial Theorem Exercise 8.1 Solutions
1. Expand the expression (1– 2x)5
Solution
By using Binomial Theorem, the expression (1-2x) can be expanded as
(1 -2x)5
= 5C0 (1)5 – 5C1 (1)4 (2x) + 5C2 (1)3 (2x)2 - 5C3 (1)2 (2x)3 + 5C4 (1)1(2x)4 + 5C5 (2x)5
= 1 – 5(2x) + 10(4x2 ) – 10(8x3 )+ 5(16x4 ) – (32x5 )
= 1 – 10x + 40x2 – 80x3 + 80x4 – 32x5
2. Expand the expression (2/x - x/2)5
Solution
By using Binomial Theorem, the expression (2/x - x/2)5 can be expanded as 
3. Expand the expression (2x – 3)6 .
Solution
By using Binomial theorem, the expression (2x - 3)6 can be expanded as
(2x - 3)6
= 6C0 (2x)6 – 6C1 (2x)5 (3) + 6C2 (2x)4 (3)2 – 6C3 (2x)3 (3)3 + 6C4 (2x)2 (3)4 – 6C5 (2x) (3)5 + 6C6 (3)6
= 64x6 – 6(32x5 )(3) + 15(16x4 )(9) – 20(8x3 )(27) + 15(4x2) (81) – 6(2x)(243) + 729
= 64x6 – 576x5 + 2160x4 – 4320x3 + 4860x2 – 2916x + 729
4. Expand the expression (x/3 + 1/x)5
Solution
By using Binomial Theorem, the expression (x/3 + 1/x)5 can be expanded as 
5. Expand (x + 1/x)6
Solution
By using Binomial Theorem, the expression(x + 1/x)6 can be expanded as 
6. 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.
It can be written that, 96 = 100 - 4
∴ (96)3 = (100 - 4)3
= 3C0 (100)3 - 3C1 (100)2 (4) + 3C2 (100)(4)2 - 3C3 (4)3
= (100)3 - 3(100)2 (4) + 3(100)(4)2 - (4)3
= 1000000 - 120000 + 4800 - 64
= 884736
7. 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)(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
= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32
= 11040808032
8. 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
= 100000000 +4000000 + 60000 + 400 + 1
= 104060401
9. Using Binomial Theorem, evaluate (99)5
Solution
99 can be written 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)5 - 5C1 (100)4 (1) + 5C2 (100)3 (1)2 - 5C3 (100)2 (1)3 + 5C4 (100)(1)4 + 5C5 (1)5
= (100)5 - 5(100)4 + 10(100)3 - 10(100)2 + 5(100) - 1
= 10000000000 - 500000000 + 10000000 - 100000 + 500 - 1
= 10010000500 - 500100001
= 9509900499
10. Using Binomial Theorem, indicate which number is larger (1.1)10000 or 1000.
Solution
By splitting 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 terms
= 1 + 10000 × 1.1 + Other positive terms
= 1 + 11000 + Other positive terms
> 1000
Hence, (1.1)10000 > 1000
11. 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 expanded as
(a + b)4 = 4C0a4 + 4C1a3b + 4C2a2b2 + 4C3ab3 + 4C4b4
(a - b)4 = 4C0a4 - 4C1a3b + 4C2a2b2 - 4C3ab3 + 4C4b4
∴ (a + b)4 - (a - b)4 = 4C0a4 + 4C1a3b + 4C2a2b2 + 4C3ab3 + 4C4b4 - [4C0a4 - 4C1a3b + 4C2a2b2 - 4C3ab3 + 4C4b4]
= 2(4C1a3b + 4C3ab3) = 2(4a3b + 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
12. Find (x + 1)6 + (x -1)6. Hence or otherwise evaluate (√2 + 1)6 + (√2 - 1)6
Solution
Using Binomial Theorem, the expressions, (x + 1)6 and (x - 1)6 , can be expanded as
(x + 1)6 = 6C0 x6 + 6C1 x5 + 6C2 x4 + 6C3 x3 + 6C4 x2 + 6C5 x + 6C6
(x - 1)6 = 6C0 x6 - 6C1 x5 + 6C2 x4 - 6C3 x3 + 6C4 x2 - 6C5 x + 6C6
∴ (x + 1)6 + (x -1)6 = 2[6C0 x6 + 6C1 x5 + 6C2 x4 + 6C4 x2 + 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
13. 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 
Thus, 9n+1 - 8n - 9 is divisible by 64, whenever n is positive integer.
14. Prove that 
Solution
By Binomial Theorem, 
By putting b = 3 and a = 1 in the above equation, we obtain 
Hence, proved.