Gujarat Board Textbook Solutions Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1
Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1 Textbook Questions and Answers.
Gujarat Board Textbook Solutions Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1
Expand each of the expressions in questions 1 – 5:
1. (1 – 2x)5
2. ((frac{2}{x}) – (frac{x}{2}))5
3. (2x – 3)6
4. ((frac{x}{3}) + (frac{1}{x}))5
5. (x + (frac{x}{1}))6
Solutions to questions 1 – 5:
1. (1 – 2x)5 = 5C0.15 + 5C1.14.(- 2x) + 5C2.13(- 2x)2
+ 5C3.12(- 2x)3 + 5C4.11(- 2x)4 + 5C5.10(- 2x)5.
= 1.1 + 5.1(- 2x) + (frac{5.4}{1.2}).1.4x2
+ (frac{5.4}{1.2}).1.(- 8x3) + (frac{5}{1}). 1.16x4 + (- 32 x5)
= 1 – 10x + 40x2 – 80x3 + 80x4 – 32x5.
2.
3. (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 + (frac{6}{1}) × (32x5)(- 3) + (frac{6.5}{1.2}).(16x4).9 + (frac{6.5.4}{1.2.3})(8x3)
(- 27) + (frac{6.5}{1.2})(4x2)(81) + (frac{6}{1})(2x)(- 243) + 729
= 64x6 – 576x5 + 2160x4 – 4320x3 + 4860x2 – 2916x + 729.
4.
5.
Using binomial theorem, evaluate each of the following questions 6 – 9:
6. (96)3
7. (102)5
8. (101)4
9. (99)5
Solutions to questions 6 – 9:
6. (96)3 = (100 – 4)3 = 3C0(100)3(- 4)0 + 3C1(100)2(- 4)
+ 3C2(100)2(- 4) + (frac{3}{1}) × 100 × 16 + (- 64)
= 1000000 – 12 × 10000 + 48 × 100 – 64
= 1000000 – 120000 + 4800 – 64
= 884736.
7. (102)5 = (100 + 2)5 = (100)5 + 5C1(100)4.2 + 5C2(100)322
+ 5C3(100)2.23+ 5C4(100)1.24 + 5C5(100)0.25
= 10000000000 +5 × (100000000) × 2 + (frac{5.4}{1.2}) × (1000000) × 4
+ (frac{5.4}{1.2}) × (10000) × 8 + (frac{5}{1}) × 100 × 16 + 32
= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32
= 11040808032.
8. (101)4 = (100 + 1)4 = 4C0(100)4 + 4C1(100)3.1
+ 4C2(100)2.12 + 4C3(100).13 + 4C4(100)0.14
= 100000000 + 4 × 1000000 + 6 × 10000 + 4 × 100 + 1
= 100000000 + 4000000 + 60000 + 400 + 1
= 104060401.
9. 995 = (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 + (- 5) × 100000000 + 10 × 1000000 – 10 × 10000 + 5 × 100 – 1
= 10000000000 – 500000000 + 10000000 – 100000 + 500 – 1
= 9509900499.
10. Using Binomial Theorem, indicate which number is larger (1.1)1000 or 1000?
Solution:
(1.1)10000 = [1 + (.1)]10000
Expanding by binomial theorem, we get
= C(10000, 0)(1)10000+ C(10000, 1)(1)10000-1(.1)
= 1 + 10000 × .1 + other terms
= 1001 + other terms of the expansion
Hence (1.1)10000 > 1000.
11. Find (a + b)4 – (a – b)4.
Hence, evaluate ((sqrt{3}) + (sqrt{2}))4 – ((sqrt{3}) – (sqrt{2}))4.
Solution:
(i) (a + b)4 = a4 + 4C1a3b + 4C2a2b2 + 4C3ab3 + 4C4b4
= a4 + 4a3b + 6a2b2 + 4ab3 + b4 ………………. (1)
(a – b)4 = a4 + 4C1a3(- b) + 4C2a2(- b)2 + 4C3a(- b)3 + 4C4(- b)4.
= a4 – 4a3b + 6a2b2 – 4ab3 + b4 …………………. (2)
Subtracting (2) from (1), e get
(a + b)4 – (a – b)4 = 2(4a3b + 4ab3) = 8ab(a2 + b2)
(ii) Putting a = (sqrt{3}), b = (sqrt{2})
∴ ((sqrt{3}) + (sqrt{2}))4 – ((sqrt{3}) – (sqrt{2}))4 = 8(sqrt{3}).(sqrt{2})[((sqrt{3}))2 + ((sqrt{2}))2
= 8(sqrt{6})(3 + 2) = 40(sqrt{6}).
12. Find (x + 1)6 + (x – 1)6. Hence, or otherwise, evaluate ((sqrt{2}) + 1)6 + ((sqrt{2}) – 1)6.
Solution:
(x + 1)6 = x6 + 6C1x5.1 + 6C2.x4.12
+ 6C3.x3.13 + 6C4.x2.14 + 6C5.x.15 + 6C6.x4.16
= x6 + 6x5 + 15x4 + 20x3 + 15x2 + 6x + 1 ……………….. (1)
(x – 1)6 = x6 + 6C1x5 × (- 1) + 6C2.x4.(- 1)2 + 6C3.x3(- 1)3
+ 6C4x2(- 1)4 + 6C5x(- 1)5 + 6C6x0(- 1)6
= x6 – 6x5 + 15x4 – 20x3 + 15x2 – 6x + 1 ………………….. (2)
Adding (1) and (2), we get
(x + 1)6 + (x – 1)6 = 2(x6 + 15x4 + 15x2 + 1)
Putting x = (sqrt{2}), we get
((sqrt{2}) + 1)6 + ((sqrt{2}) – 1)6
= 2[((sqrt{2}))6 + 15((sqrt{2}))4 + 15((sqrt{2}))2 + 1]
= 2[8 + 60 + 30 + 1] = 2 × 99
= 198.
13. Show that 9n+1 +1 – 8n – 9 is divisible by 64, whenever n is a positive integer?
Solution:
Writing the binomial expansion of (1 + x)n+1, we have:
(1 + x)n+1 = C(n + 1, 0) + C(n + 1, 1)x
+ C(n + 1, 2)x2 + C(n + 1, 3)x3 + … + 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)(8)2
C(n + 1, 3)83 + … + C(n + 1, n + 1)8n+1
or which is the required expansion of (1 + x)n+1, when x = 8.
or (9)n+1 = 1 + (n + 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 (9)n+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)84 + ……. + C(n + 1, n + 1)8n+1
= 64 × some constant quantity.
Hence, 9n+1 + 1 – 8n – 9 is divisible by 64, whenever re is a positive integer (Proved).
14. Prove that (sum_{r=0}^{n}) 3r nCr = 4n.
Solution:
(sum_{r=0}^{n}) 3r nCr = nC0 + nC131
+ nC2.32 + ……………. + nC3.3r + ………………… + nCn3n.
= (1 + 3)4 = 4n.