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

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

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

Answer

(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


Answer

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

3. (2x – 3)6

Answer

(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

Answer

(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

Answer

(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

Answer

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

Answer

(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

Answer

(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

Answer

(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

Answer

(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

Answer

(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.

Answer

(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

Answer

(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

Answer

(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)
Adding (i) and (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.

Answer

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

Answer


3r nCr = nC0 + nC1.31 + nC232 + ………….. + nCr.3r + ……….. + nCn.3n = (1 + 3)n = 4n.
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