## NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Exercise 8.1

If you want NCERT Solutions of Chapter 8 Binomial Theorem Exercise 8.1 then you can get here. We have provided Class 11 Maths NCERT Solutions that is prepared by Studyrankers which are detailed and accurate. These Solutions are updated according to latest pattern released by CBSE.

1. Expand each of the expressions in Exercises 1 – 5.
(1 – 2x) 5

(1 – 2x) 5
5C0 .15 + 5C1 .14. (– 2x) + 5C2 .13 (– 2x)2 + 5C3 .12 (– 2x) 3 + 5C4 .11 (– 2x) 4 + 5C0 (– 2x)5
= 1.1 + 5.1. (– 2x) + (5.4 )/(1.2) .1. 4x 2 + (5.4)/( 1.2) .1 (–8x 3) + 5 /1 .1.16 x4 + (-32 x5)
= 1 – 10x + 40x 2 – 80x 3 + 80x – 32x 5.

2. (2/x – x/2)5

= 32x-5 – 40x-3 + 20x-1 – 5x + 5/8 x3 – 1/32 x5

3. (2x – 3)6

(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(2x)0 (-3)6
= 64x6 + 6/1 (32x5)(-3) + ((6 . 5)/(1 . 2)) (16x4)9 + ((6 . 5 . 4)/(1 . 2 . 3))(8x3)(-27) + ((6 . 5)/(1 . 2)) (4x2) 81 + 6/1 (2x)(-243) + 729
= 64x6 – 576x5 + 2160x4 – 4320x3 + 4860x2 – 2916x + 729

4. (x/3 + 1/x)5

(x/3 + 1/x)5 = 5C0(x/3)5(1/x)0 + 5C1(x/3)4(1/x) + 5C2(x/3)3(1/x)2 + 5C3(x/3)2(1/x)3 + 5C4(x/3)(1/x)4 + 5C5(x/3)0(1/x)5
= x5/243 + 5/1 . x4/81 . 1/x + (5.4)/(1.2) . x3/27 .1/x2 + (5.4)/(1.2) . x2/9 . 1/x3 + 5/1 . x/3 . 1/x4 + 1/x5
= x5/243 + 5/81 . x3 + 10/27 . x + 10/9 . 1/x + 5/3 . 1/x3 + 1/x5

5. (x + 1/x)6

(x + 1/x)6 = 6C0x6(1/x)0 + 6C1x5(1/x) + 6C2x4(1/x)2 + 6C3x3(1/x)3 + 6C4x2(1/x)4 + 6C5x(1/x)5 + 6C6x0(1/x)6
= x6 + 6/1 . x4 + ((6 . 5)/(1 . 2))x2 + ((6 . 5. 4)/(1 . 2 . 3)) + (6 . 5)/(1 . 2) (1/x2) + 6/1(1/x4) + (1/x6)
= x6 + 6x4 + 15x2 + 20 + 15 1/x2 + 6/x4 + 1/x6

6. Using Binomial Theorem, evaluate the following:
(96)3

We express 96 as the sum or difference of two numbers whose powers are easier to calculate, and then use Binomial Theorem.
Write 96 = 100 – 4
Therefore
(96)3 = (100 – 4)3
3C0 (100)3 – 3C1 (100)2 (4) + 3C2 (100)(4)2 – 3C3 (4)3
= 1000000 – 3(10000) (4) + 3 (100) (16) – (64)
= 1000000 – 120000 + 4800 – 64 = 884736

7. Using binomial theorem, evaluate the value of (102)5

(102)5 = (100 + 2)5
= 1005 + 5C1(100)42 + 5C2(100)322 + 5C3(100)223 + 5C4(100)24 + 5C5(100)025
= 10000000000 + 5 × (100000000) × 2 + ((5 . 4)/(1 . 2))(1000000) × 4 + ((5 .4)/(1 . 2))(10000) × 8 + 5/1 (100) × 16 + 32
= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32
= 11040808032

8. Using Binomial Theorem, evaluate the following: (101)4

(101)4 = (100 + 1)4
4C0 (100)4 + 4C1 (100)3 (1) + 4C2 (100)2 (1)2 + 4C3 (100)1 (1)3 + 4C4 (1)4
= 100000000 + 4 (1000000) + 6(10000) + 4 (100) + 1
= 100000000 + 4 000000 + 60000 + 4 00 + 1
= 104060401

7. Using binomial theorem, evaluate the value of (102)5

(102)5 = (100 + 2)5
= 1005 + 5C1(100)42 + 5C2(100)322 + 5C3(100)223 + 5C4(100)24 + 5C5(100)025
= 10000000000 + 5 × (100000000) × 2 + ((5 . 4)/(1 . 2))(1000000) × 4 + ((5 .4)/(1 . 2))(10000) × 8 + 5/1 (100) × 16 + 32
= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32
= 11040808032

8. Using Binomial Theorem, evaluate the following: (101)4

(101)4 = (100 + 1)4
4C0 (100)4 + 4C1 (100)3 (1) + 4C2 (100)2 (1)2 + 4C3 (100)1 (1)3 + 4C4 (1)4
= 100000000 + 4 (1000000) + 6(10000) + 4 (100) + 1
= 100000000 + 4 000000 + 60000 + 4 00 + 1
= 104060401

9. Using Binomial theorem evaluate (99)5

(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 + (–1)5
= 10000000000 – 500000000 + 10000000 – 100000 + 500 – 1
= 9509900499

10. Using Binomial theorem indicate which number is larger (1.1)10000 or 1000.

(1.1)10000 = [1 + (0.1)]10000
Expanding by binomial theorem
= C(10000,0) (1)10000 + C(10000, 1) (1)10000 – 1(0.1) + other terms
= 1 + 10000 × 0.1 + other terms
= 1001 + other terms
Hence, (1.1)10000 > 1000.

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

(a + b )4 = a4 + 4C1a3b + 4C2a2b2 + 4C3a1b3 + 4C4a0b4
= a4 + 4a3b + 6a2b2 + 4ab3 + b4 ……(i)
(a – b)4 = a4 + 4C1a3(-b) + 4C2a2(-b)2 + 4C3a(-b)3 + 4C4(-b)4
= a4 – 4a3b + 6a2b2 – 4ab3 + b4 ……….(ii)
Subtracting (ii) from (i)
(a + b)4 – (a – b)4 = 2[4a3b + 4ab3]
= 8ab(a2 + b2) …….(iii)
Putting a = √3, b = √2 in equ. (iii)
(√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

(x + 1)6 = x6 + 6C1.x5.1 + 6C2. x4. 12 + 6C3.x3.13 + 6C4.x2.14 + 6C5.x.15 + 6C6.x0.16
= x6 + 6x5 + 15x4 + 20x3 + 15x2 + 6x + 1 ….(i)
(x -1)6 = x6 + 6C1.x5.(-1) + 6C2.x4.(-1)6C3.x3.(-1)3 + 6C4.x2.(-1)4 + 6C5.x.(-1)5 + 6C6.x0.(-1)6
= x6 – 6x5 + 15x4 – 20x3 + 15x2 – 6x + 1 …(ii)
(x + 1)6 + (x – 1)6 = 2[x6 + 15x4 + 15x2 + 1]
Putting x = √2
(√2 + 1)6 + (√2 – 1)6
= 2[(√2)6 + 15(√2)4 + 15(√2)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.

We have
(1 + x) n + 1 = C(n + 1, 0) + C(n + 1, 1)x + C(n + 1, 2) x + C(n + 1, 3) x 3 + ...........+ C(n + 1, n + 1) xn + 1 Putting x = 8, we get
(1 + 8)n + 1 = C(n + 1, 0) + C(n + 1, 1)8 + C(n + 1, 2) 82 + C(n + 1, 3) 83 + ........... + C(n + 1, n + 1) 8n + 1
9n + 1 = C(n +1, 0) + C(n + 1, 1)8 + C(n + 1, 2) 82 + C(n + 1, 3) 83 + ........... + C(n + 1, n + 1) 8n + 1
[∵ C(n + 1, 0) = 1 and C(n + 1, 1) = n + 1]
or 9n + 1 = 1 + 8n + 8 + C(n + 1, 2) 82 + C(n + 1, 3) 83 + ........... + C(n + 1, n + 1) 8n + 1
or 9n + 1 – 8n – 9
= C(n + 1, 2) 82 + C(n + 1, 3) 83 + ........... + C(n + 1, n + 1) 8n + 1
= 82 [C(n + 1, 2) + C(n + 1, 3) 8 + C(n + 1, 4) 82 + ........... + C(n + 1, n + 1) 8n – 1]
9n + 1 – 8n – 9 = 64 × some constant quantity.
Hence, 9n + 1 – 8n – 9 is divisible by 64 whenever n is a positive integer.

14. Prove that 3r nCr = 4n