CLASS 11 MATHS CHAPTER-8 BINOMIAL THEOREM 

Exercise 8.1

Expand each of the expression in Exercises 1 to 5.

Question1. Expand the expression (1– 2x)⁵

Solution :
Using Binomial Theorem,

Question2. 

Solution :
Using Binomial Theorem,

Question3. Expand the expression (2x – 3)⁶ 

Solution :
Using Binomial Theorem,

Question4. 

Solution :
Using Binomial Theorem,

Question5. 

Solution :
Using Binomial Theorem,

Using binomial theorem evaluate each of the following:

Question6. Using Binomial Theorem, evaluate (96)³

Solution :
Given: 96³

 ^{96 = 100 – 4}

Using Binomial Theorem,

(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
 

Question 7 .Using Binomial Theorem, evaluate (102)⁵

Solution :
Given: (102)⁵

Using Binomial Theorem,

= (102)⁵  = (100+2)⁵ = ⁵C₀(100)⁵ + ⁵C₁(100)⁴ (2) + ⁵C₂ (100)³(2)² + ⁵C₃(100)²(2)³ + ⁵ C₄(100)(2)⁴ + ⁵C₅(2)⁵

=10000000000+5(100000000)(2)+10(1000000)(4)+10(10000)(8)+5(100)(16)+32

= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32

= 11040808032

Question8. Using Binomial Theorem, evaluate (101)⁴

Solution :
Given: (101)⁴

Using Binomial Theorem,

101 = 100+1

 

(101)⁴=(100+1)⁴

=⁴C₀(100)⁴+⁴C₁(100)³(1)+⁴C₂(100)²(1)²+⁴C_{^{3 (}}100)(1)³+⁴C₄(1)⁴

=100000000+4(1000000)+6(10000)+4(100)+1

= 100000000 + 4000000 + 60000 + 400 + 1

= 104060401

9. Using Binomial Theorem, evaluate (99)⁵

Solution :
Given: (99)⁵

Using Binomial Theorem,

(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

=10000000000−5(100000000)−10(1000000)−10(10000)+5(100)−1=

= 10000000000 – 500000000 + 10000000 – 100000 + 500 – 1

= 9509900499

Question10. Using Binomial Theorem, indicate which number is larger (1.1)¹⁰⁰⁰⁰ or 1000.

Solution :
Given:  (1.1)¹⁰⁰⁰⁰ or 1000.

Using Binomial Theorem,

By splitting 1.1 and then applying Binomial Theorem, the first few terms of (1.1)10000

 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

Question11. Find (a + b)⁴ – (a – b)⁴. Hence, evaluate(√3 + √2)⁴– (√3 – √2)^4.

Solution :
Given: (a + b)⁴ – (a – b)⁴

Using Binomial Theorem,

Question12. Find (x + 1)⁴ – (x – 1)⁴  Hence or otherwise evaluate (√ 2 + 1 )⁶ + (6 -1 )⁶

Solution :
Given: (x + 1)⁴ – (x – 1)⁴

Using Binomial Theorem,

Question13. Show that 9 ⁿ⁺¹ is divisible by 64 whenever n is a positive integer.

Solution :
Given:   9 ⁿ⁺¹ is divisible by 64

 In order to show that 9ⁿ⁺¹ −8n−9 is divisible by 64, it has to be prove that, 9ⁿ⁺¹−8n−9=64k

, where k is some natural number.

Using Binomial Theorem,

Question14. Prove that 

Solution :
 

Exercise 8.2

Find the coefficient of:

Question1. Find the coefficient of x⁵ in (x + 3)⁸

Solution :It is known that (r+1)^th  term, (T_r+1). in the binomial expansion of (a+b)ⁿ  is given by T_r+1 = ⁿC_ra^n−r b^r

Assuming that x²
occurs in the (r+1)ⁿ term of the expansion (x+3)⁸, we obtain T_^r+1 = ^aC_r(x)^8-r(3)^r

Comparing the indices of x
 in x5 in T_r+1. We obtain r = 3 Thus, the coefficient of x⁵  is ⁸C₃(3)³=8! / 3!5!×3³

=(8⋅7⋅6⋅5!  /  3⋅2.5!)∗3³=1512

 

Question2. Find the coefficient of a⁵b⁷ in (a – 2b)¹²

Solution :
General form of the expansion ( r + 1)^th is (T _r+2)………(i)

Write the general term in the expansion of

Question3. Write the general term in the expansion of (x² – y)⁶

Solution :
 It is known that the general term T_r+1{ which is the (r+1)ⁿ term } in the binomial expansion of (a+b)n is given by T_r+1=ⁿC_ra^n-rb^r.

Thus, the general term in the expansion of (x²−y⁶)
 is T_r+1 = ⁶C_r(x²)^6−r (−y)^r = (−1)^r6 C_r – x^{12 −2r}⋅y^r

Question4. Write the general term in the expansion of (x² – yx)¹², x ≠ 0

Solution :
General form of the expansion T_r+1 is

{ which is the (r+1)th term } in the binomial expansion of (a+b)n is given by Tr+1=nCran−rbr.

Thus, the general term in the expansion of (x2−yx)12
 is T_r+1 = ¹²C_r(x²)^12−t(−yx)^r

=(−1)^r12C_r−x^24−2r−y^r

=(−1)^r−2C_rx^24−r⋅y^r

Question5. Find the 4^th term in the expansion of (x – 2y)¹² .

Solution :
It is known (r+1)ⁿ term, T_r+1. in the binomial expansion of (a+b)ⁿ

Question6. Find the 13^th term in the expansion of

Solution :
It is known (r+1)^th term, T_r+1, in the binomial expansion of (a+b)ⁿ is given by T_ε+1 = ⁿC_raⁿ⁻¹b^r

Thus, the 13^th
 term in the expansion of

Find the middle terms in the expansion of:

Question7. 

Solution :
 

Question8. (x/3 + 9y)¹⁰

Solution :
Here (a + b )ⁿ which is an even number.

Therefore, the middle terms are (x/3 + 9y)¹⁰ is 6th term.

General form of the expansion

Question9. In the expansion of (1 + a)^{m + n}, prove that coefficients of am and an are equal.

Solution :
 

Question10. The coefficients of the (r – 1)^th, r^th and (r + 1)^th terms in the expansion of (x + 1)ⁿ are in the ratio 1:3:5. Find n and r.

Solution :
 

Question11.  Prove that the coefficient of xn in the expansion of (1 + x)²ⁿ is twice the coefficient of xn in the expansion of (1 + x)^2n–1 .

Solution :
 

= It is known that (r+1)^th  term, (T_r+1), in the binomial expansion of (a+b)ⁿ is given by T_r+1=ⁿC_ra^n−rb^r

Assuming that xⁿ
 occurs in the (r+1)ⁿ term of the expansion of (1+x)²ⁿ, we obtain T_r+1=²ⁿCr(1)^2n−r(x)^r

=²ⁿC_r(x)^r

Comparing the indices of x
 in xn and in T_r+1, we obtain r=n Therefore, the coefficient of xn in the expansion of (1+x)²ⁿ is

Miscellaneous Exercise

Question1. Find a, b and n in the expansion of (a + b)ⁿ if the first three terms of the expansion are 729, 7290 and 30375, respectively.

Solution :
Given: (a + b)ⁿ

T_r+1 = ⁿC_r a^n-t b^r

The first three terms of the expansion are given as 729, 7290 and 30375 respectively. Then we have,

T1 = ⁿC₀ a^n-0 b⁰ = an = 729….. 1

T2 =ⁿC₁ a^n-1 b¹ = ⁿa^n-1 b = 7290…. 2

T3 = ⁿC₂ a^n-2 b² = {n (n -1)/2 }a^n-2 b² = 30375……3

Dividing 2 by 1 we get

Question2. Find a if the coefficients of x² and x³ in the expansion of (3 + ax)⁹ are equal.

Solution :

Question3. Find the coefficient of x⁵ in the product (1 + 2x)⁶ (1 – x)⁷ using binomial theorem.

Solution :
Using Binomial Theorem,

Question4. 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]

Solution :
 

n 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.

a can be written as a = a – b + b

an = (a – b + b)ⁿ = [(a – b) + b]ⁿ

= ⁿC₀ (a – b)ⁿ + ⁿC₁ (a – b)^n-1 b + …… + ⁿ C _n bⁿ

aⁿ – bⁿ = (a – b) [(a –b)^n-1 + ⁿC₁ (a – b)^n-1 b + …… + ⁿ C _n bⁿ]

aⁿ – bⁿ = (a – b) ^k

Where k = [(a –b)^n-1 + ⁿC₁ (a – b)^n-1 b + …… + ⁿ C _n b^n] is a natural number

 

Question5. Evaluate: (√3 + √2)⁶ – (√3 – √2)⁶

Solution :
 

Question6. Find the value of

Solution :

Question7. Find an approximation of (0.99)⁵ using the first three terms of its expansion.

Solution :
Here .99 can be written as

0.99 = 1 – 0.01

Now by applying binomial theorem we get

(o. 99)5 = (1 – 0.01)⁵

= 5C0 (1)⁵ – ⁵C₁ (1)⁴ (0.01) + ⁵C₂ (1)³ (0.01)²

= 1 – 5 (0.01) + 10 (0.01)²

= 1 – 0.05 + 0.001

= 0.951

Question8. Find n if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of is 

Solution :

Question9. Expand using binomial theorem 

Solution :
We have

Question10. Find the expansion of (3x² – 2ax + 3a²)³using binomial theorem.

Solution :
Here

=  We know that (a + b)³ = a³ + 3a²b + 3ab² + b³

Putting a = 3x² & b = -a (2x-3a), we get
[3x² + (-a (2x-3a))]³

= (3×2)³+3(3x²)²(-a (2x-3a)) + 3(3x²) (-a (2x-3a))² + (-a (2x-3a))³

= 27x⁶ – 27ax⁴ (2x-3a) + 9a2x² (2x-3a)² – a³(2x-3a)³

= 27x⁶ – 54ax⁵ + 81a2x⁴ + 9a2x² (4x²-12ax+9a²) – a3 [(2x)3 – (3a)3 – 3(2x)²(3a) + 3(2x)(3a)²]

= 27x⁶ – 54ax⁵ + 81a²x⁴ + 36a²x⁴ – 108a³x³ + 81a⁴x² – 8a3x3 + 27a⁶ + 36a⁴x2 – 54a⁵x

= 2^7×6 – 54ax⁵+ 117a²x⁴ – 116a³x³ + 117a⁴x² – 54a⁵x + 27a⁶

Thus, (3x² – 2ax + 3a²)³

= 27x⁶ – 54ax⁵+ 117a2x⁴ – 116a³x³ + 117a⁴x² – 54a⁵x + 27a⁶