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NCERT Class 11 Mathematics Chapter 7 Binomial Theorem
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Binomial Theorem
Chapter – 7
| Exercise 7.1 |
Expand each of the expressions in Exercises 1 to 5
1. (1 – 2x)⁵
Ans: By using Binomial Theorem, the expression (1 – 2x)⁵ can be expanded as
(1 – 2x)⁵
= ⁵C₀ (1)⁵ – ⁵C₁ (1)⁴ (2x) + ⁵C₂ (1)³ (2x)² – ⁵C₃ (l)² (2x)³ + ⁵C₄ (l)¹ (2x)⁴ – ⁵C₅ (2x)⁵
= 1 – 5(2x) + 10(4x²) – 10(8x³) + 5(16x⁴) – (32x⁵)
= 1 – 10x + 40x² – 80x³ + 80x⁴ – 32x⁵
Ans:
3. (2x – 3)⁶.
Ans: By using Binomial Theorem, the expression ( 2 x – 3)⁶ can be expanded as
(2x – 3)⁶ = ⁶C₀ (2x)⁶ – ⁶C₁ (2x)³ (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
Ans:
Ans:
Using binomial theorem, evaluate each of the following:
6. (96)³.
Ans: 96 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, a 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. (102)⁵.
Ans: 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. (101)⁴.
Ans: 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
= 10000000 + 4000000 + 60000 + 400 + 1
= 104060401
9. (99)⁵.
Ans: 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 (l)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.
Ans: 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.
Ans:
Putting a = √3 and b = √2
(√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.
Ans:
13. Show that 9ⁿ⁺¹ – 8n – 9 is divisible by 64, whenever n is a a positive integer.
Ans: 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 + ᵐC₁a + ᵐC₂a² +…+ ᵐCₘ aᵐ
For a = 8 and m = n + 1, we obtain
Ans:
| Miscellaneous Exercise on Chapter 7 |
1. If a and b are distinct integers, prove that a – b is a factor of aⁿ – bⁿ , whenever n is a positive integer.
[Hint write aⁿ = (a – b + b)ⁿ and expand]
Ans: In order to prove that (a – b) is a factor of ( aⁿ – bⁿ), it has to be proved that
aⁿ – bⁿ = k(a – b), where k is some natural number
It can be written that, a = a a – b + b
∴ aⁿ =(a-b+b)ⁿ = [(a – b) + b]ⁿ
This shows that (a – b) is a factor of (aⁿ – b”), where n is a positive integer.
2. Evaluate (√3 + √2)⁶ – (√3 – √2)⁶.
Ans: Firstly, the expression (a + b)⁶ – (a – b)⁶ is simplified by using Binomial Theorem. This can be done as
(a+b)⁶ = ⁶C₀ a⁶ + ⁶C₁a⁵b + ⁶C₂a⁴ b² + ⁶C₃a³b³ + ⁶C₄ a² b⁴ + ⁶C₅ a¹ b⁵ + ⁶C₆ b₆
= a⁶ + 6a⁵ b + 15a⁴ b² + 20a³ b³ + 15a² b⁴ + 6ab⁵ + b⁶
(a-b)⁶ = ⁶C₀ a⁶ – ⁶C₁ a⁵ b + ⁶C₂ a⁴ b² – ⁶C₃ a³ b³+ ⁶C₄ a² b⁴ – ⁶C₅ a¹ b⁵ + ⁶C₆ b₆
= a⁶ – 6a⁵ b + 15a⁴ b² – 20a³ b³ + 15a² b⁴ – 6ab⁵ + b⁶
∴ (a + b)⁶ – (a – b)⁶ = 2[6a⁵ b + 20a³ b³ + 6ab⁵]
Putting a = √3 and b = √2, we obtain
(√3 + √2)⁶ – (√3 – √2)⁶ = 2[6(√3)⁵ (√2) + 20(√3)³ (√2)³ + 6(√3) (√2)⁵]
= 2[54√6 + 120√6 + 24√6]
= 2 × 198√6
= 396√6
Ans: Firstly, the expression (x + y)⁴ + (x – y)⁴ is simplified by using Binomial Theorem.
This can be done as
(x+y)⁴ = ⁴C₀ x⁴ + ⁴C₁ x³y + ⁴C₂ x²y² + ⁴C₃xy³ + ⁴C₄ y⁴
= x⁴+ 4x³ y + 6x²y² + 4xy³ + y⁴
(x-y)⁴ = ⁴C₀ x⁴ – ⁴C₁ x³y + ⁴C₂ x² y² – ⁴C₃ xy³+ ⁴C₄ y⁴
= x⁴ – 4x ³ y + 6x²y²- 4xy³ + y⁴
∴(x + y)⁴ + (x – y)⁴ = 2(x⁴ + 6x²y² + y⁴)
Putting x = a² and y = √a² – 1, we obtain
( a² + √a² -1 )⁴ + (a² – √a ² – 1)⁴ = 2[(a²)⁴ + 6(a²)² (√a² – 1)² + (√a² – 1)]
= 2[a⁸ + 6a⁴ (a² – 1) + (a² – 1)² ]
= 2[a⁸+ 6a⁶ – 6a⁴ + a⁴ – 2a² + 1]
= 2[a⁸ + 6a⁶ – 5a⁴ – 2a² + 1]
= 2a⁸ + 12a⁶ – 10a⁴ – 4a² + 2
4. Find an approximation of (0.99)⁵ using the first three terms of its expansion.
Ans: 0.99 =1 – 0.01
∴ (0.99)⁵ = (1 – 0.01)⁵
= ⁵C₀ (1)⁵ – ⁵C₁ (1)⁴ (0.01)+ ⁵C₂ (1)³ (0.01)²
(Approximately)
= 1 – 5(0.01) + 10(0.01)²
= 1 – 0.05+0.001
=1.001-0.05
= 0.951
Thus, the value of (0.99)⁵ is approximately 0.951.
Ans:
6. Find the expansion of (3x² – 2ax + 3a²)³ using the binomial theorem.
Ans: Using Binomial Theorem, the given expression [(3x² – 2ax + 3a²]³can be expanded as
[(3x² – 2ax)+ 3a²]³
= ³C₀ (3x² – 2ax)³ + ³C₁ (3x² – 2ax)² (3a²) + ³C₂(3x² -2ax)(3a²)² + ³C₃ (3a²)³
= (3x² – 2ax)³ + 3(9x⁴ – 12ax³ + 4a²x²)(3a²) + 3(3x² – 2ax)(9a⁴) + 27a⁶
= (3x² – 2ax)³ + 81a²x⁴ – 108a³x³ + 36a⁴ x² + 81a⁴ x² – 54a⁵ x + 27a⁶
= (3x² – 2ax)³ + 81a² x⁴ – 108a³ x³ + 117a⁴x²- 54a⁵ x + 27a⁶ ……..(1)
Again by using Binomial Theorem, we obtain
(3x² – 2ax)³
= ³C₉ (3x²)³ – ³C₁ (3x²)² (2ax) + ³C₂ (3x²)(2ax)² – ³C₃ (2ax)³
= 27x⁶ – 3(9x⁴)(2ax) + 3(3x²)(4a² x²) – 8a³ x³
= 27x⁶ – 54ax⁵ + 36a² x⁴ – 8a³x³ ……..(2)
From (1) and (2), we obtain
(3x² – 2ax + 3a²)³
= 27x⁶ – 54ax⁵+ 36a² x⁴ – 8a³x³ + 81a²x⁴ – 108a³x³ + 117a⁴x² – 54a⁵ x + 27a⁶
= 27x⁶ – 54ax⁵ + 117a²x⁴ – 116a³x³ + 117a⁴x² – 54a²x +27a⁶
