# NCERT Solutions for Class 11 Maths Chapter 8- Binomial Theorem

*According to the Revised CBSE Syllabus 2020-21, this chapter has been removed.

The NCERT Solutions Class 11 Chapter 8 Binomial Theorem can be downloaded at BYJU’S without any hassle. Practising these solutions can help the students clear their doubts as well as to solve the problems faster. Students can learn new tricks to answer a particular question in different ways giving them an edge with the exam preparation.

The concepts covered in Chapter 8 of the Maths textbook includes the study of essential topics such as Positive Integral Indices, Pascal’s Triangle, Binomial theorem for any positive integer and some special cases. Students can score high marks in the exams with ease by practising the NCERT Solutions for all the questions present in the textbook. Each solution is mounted step-by-step, considering the perception level of the students. So, it gets pretty apparent to understand the logic set behind each answer and develop a better comprehension of the concepts.

### Access Answers to NCERT Class 11 Maths Chapter 8

Exercise 8.1 Page No: 166

Expand each of the expressions in Exercises 1 to 5.

1. (1 – 2x)5

Solution:

From binomial theorem expansion we can write as

(1 – 2x)5

= 5Co (1)55C1 (1)4 (2x) + 5C2 (1)3 (2x)25C3 (1)2 (2x)3 + 5C4 (1)1 (2x)45C5 (2x)5

= 1 – 5 (2x) + 10 (4x)2 – 10 (8x3) + 5 ( 16 x4) – (32 x5)

= 1 – 10x + 40x2 – 80x3 – 32x5

Solution:

From binomial theorem, given equation can be expanded as

3. (2x – 3)6

Solution:

From binomial theorem, given equation can be expanded as

Solution:

From binomial theorem, given equation can be expanded as

Solution:

From binomial theorem, given equation can be expanded as

6. (96)3

Solution:

Given (96)3

96 can be expressed as the sum or difference of two numbers and then binomial theorem can be applied.

The given question can be written as 96 = 100 – 4

(96)3 = (100 – 4)3

= 3C0 (100)33C1 (100)2 (4) – 3C2 (100) (4)23C3 (4)3

= (100)3 – 3 (100)2 (4) + 3 (100) (4)2 – (4)3

= 1000000 – 120000 + 4800 – 64

= 884736

7. (102)5

Solution:

Given (102)5

102 can be expressed as the sum or difference of two numbers and then binomial theorem can be applied.

The given question can be written as 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 + 5 (100) (2)3 + 5 (100) (2)4 + (2)5

= 1000000000 + 1000000000 + 40000000 + 80000 + 8000 + 32

= 11040808032

8. (101)4

Solution:

Given (101)4

101 can be expressed as the sum or difference of two numbers and then binomial theorem can be applied.

The given question can be written as 101 = 100 + 1

(101)4 = (100 + 1)4

= 4C0 (100)4 + 4C1 (100)3 (1) + 4C2 (100)2 (1)2 + 4C3 (100) (1)2 + 4C4 (1)4

= (100)4 + 4 (100)3 + 6 (100)2 + 4 (100) + (1)4

= 100000000 + 400000 + 60000 + 400 + 1

= 1040604001

9. (99)5

Solution:

Given (99)5

99 can be written as the sum or difference of two numbers then binomial theorem can be applied.

The given question can be written as 99 = 100 -1

(99)5 = (100 – 1)5

= 5C0 (100)55C1 (100)4 (1) + 5C2 (100)3 (1)25C3 (100)2 (1)3 + 5C4 (100) (1)45C5 (1)5

= (100)5 – 5 (100)4 + 10 (100)3 – 10 (100)2 + 5 (100) – 1

= 1000000000 – 5000000000 + 10000000 – 100000 + 500 – 1

= 9509900499

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

Solution:

By splitting the given 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

= (1 + 0.1)10000 C1 (1.1) + other positive terms

= 1 + 10000 × 1.1 + other positive terms

= 1 + 11000 + other positive terms

> 1000

(1.1)10000 > 1000

11. Find (a + b)4 – (a – b)4. Hence, evaluate

Solution:

Using binomial theorem the expression (a + b)4 and (a – b)4, can be expanded

(a + b)4 = 4C0 a4 + 4C1 a3 b + 4C2 a2 b2 + 4C3 a b3 + 4C4 b4

(a – b)4 = 4C0 a44C1 a3 b + 4C2 a2 b24C3 a b3 + 4C4 b4

Now (a + b)4 – (a – b)4 = 4C0 a4 + 4C1 a3 b + 4C2 a2 b2 + 4C3 a b3 + 4C4 b4 – [4C0 a44C1 a3 b + 4C2 a2 b24C3 a b3 + 4C4 b4]

= 2 (4C1 a3 b + 4C3 a b3)

= 2 (4a3 b + 4ab3)

= 8ab (a2 + b2)

Now by substituting a = √3 and b = √2 we get

(√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

Solution:

Using binomial theorem the expressions, (x + 1)6 and (x – 1)6 can be expressed as

(x + 1)6 = 6C0 x6 + 6C1 x5 + 6C2 x4 + 6C3 x3 + 6C4 x2 + 6C5 x + 6C6

(x – 1)6 = 6C0 x66C1 x5 + 6C2 x46C3 x3 + 6C4 x26C5 x + 6C6

Now, (x + 1)6 – (x – 1)6 = 6C0 x6 + 6C1 x5 + 6C2 x4 + 6C3 x3 + 6C4 x2 + 6C5 x + 6C6 – [6C0 x66C1 x5 + 6C2 x46C3 x3 + 6C4 x26C5 x + 6C6]

= 2 [6C0 x6 + 6C2 x4 + 6C4 x2 + 6C6]

= 2 [x6 + 15x4 + 15x2 + 1]

Now by substituting x = √2 we get

(√2 + 1)6 – (√2 – 1)6 = 2 [(√2)6 + 15(√2)4 + 15(√2)2 + 1]

= 2 (8 + 15 × 4 + 15 × 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.

Solution:

In order to show that 9n+1 – 8n – 9 is divisible by 64, it has to be show that 9n+1 – 8n – 9 = 64 k, where k is some natural number

Using binomial theorem,

(1 + a)m = mC0 + mC1 a + mC2 a2 + …. + m C m am

For a = 8 and m = n + 1 we get

(1 + 8)n+1 = n+1C0 + n+1C1 (8) + n+1C2 (8)2 + …. + n+1 C n+1 (8)n+1

9n+1 = 1 + (n + 1) 8 + 82 [n+1C2 + n+1C3 (8) + …. + n+1 C n+1 (8)n-1]

9n+1 = 9 + 8n + 64 [n+1C2 + n+1C3 (8) + …. + n+1 C n+1 (8)n-1]

9n+1 – 8n – 9 = 64 k

Where k = [n+1C2 + n+1C3 (8) + …. + n+1 C n+1 (8)n-1] is a natural number

Thus, 9n+1 – 8n – 9 is divisible by 64, whenever n is positive integer.

Hence the proof

14. Prove that

Solution:

Exercise 8.2 Page No: 171

Find the coefficient of

1. x5 in (x + 3)8

Solution:

The general term Tr+1 in the binomial expansion is given by Tr+1 = n C r an-r br

Here x5 is the Tr+1 term so a= x, b = 3 and n =8

Tr+1 = 8Cr x8-r 3r…………… (i)

For finding out x5

We have to equate x5= x8-r

⇒ r= 3

Putting value of r in (i) we get

= 1512 x5

Hence the coefficient of x5= 1512

2. a5b7 in (a – 2b)12 .

Solution:

The general term Tr+1 in the binomial expansion is given by Tr+1 = n C r an-r br

Here a = a, b = -2b & n =12

Substituting the values, we get

Tr+1 = 12Cr a12-r (-2b)r………. (i)

To find a5

We equate a12-r =a5

r = 7

Putting r = 7 in (i)

T8 = 12C7 a5 (-2b)7

= -101376 a5 b7

Hence the coefficient of a5b7= -101376

Write the general term in the expansion of

3. (x2 – y)6

Solution:

The general term Tr+1 in the binomial expansion is given by

Tr+1 = n C r an-r br…….. (i)

Here a = x2 , n = 6 and b = -y

Putting values in (i)

Tr+1 = 6Cr x 2(6-r) (-1)r yr

= -1r 6cr .x12 – 2r. yr

4. (x2 – y x)12, x ≠ 0.

Solution:

The general term Tr+1 in the binomial expansion is given by Tr+1 = n C r an-r br

Here n = 12, a= x2 and b = -y x

Substituting the values we get

Tn+1 =12Cr × x2(12-r) (-1)r yr xr

= -1r 12cr .x24 –2r. yr

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

Solution:

The general term Tr+1 in the binomial expansion is given by Tr+1 = n C r an-r br

Here a= x, n =12, r= 3 and b = -2y

By substituting the values we get

T4 = 12C3 x9 (-2y)3

= -1760 x9 y3

6. Find the 13th term in the expansion of

Solution:

Find the middle terms in the expansions of

Solution:

Solution:

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

Solution:

We know that the general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br

Here n= m+n, a = 1 and b= a

Substituting the values in the general form

Tr+1 = m+n Cr 1m+n-r ar

m+n Cr ar…………. (i)

Now we have that the general term for the expression is,

Tr+1 =  m+n Cr ar

Now, For coefficient of am

Tm+1 =  m+n Cm am

Hence, for coefficient of am, value of r = m

So, the coefficient is m+n C m

Similarly, Coefficient of an is m+n C n

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

Solution:

The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br

Here the binomial is (1+x)n with a = 1 , b = x and n = n

The (r+1)th term is given by

T(r+1) = nCr 1n-r xr

T(r+1) = nCr xr

The coefficient of (r+1)th term is nCr

The rth term is given by (r-1)th term

T(r+1-1) = nCr-1 xr-1

Tr = nCr-1 xr-1

∴ the coefficient of rth term is nCr-1

For (r-1)th term we will take (r-2)th term

Tr-2+1 = nCr-2 xr-2

Tr-1 = nCr-2 xr-2

∴ the coefficient of (r-1)th term is nCr-2

Given that the coefficient of (r-1)th, rth and r+1th term are in ratio 1:3:5

Therefore,

⇒ 5r = 3n – 3r + 3

⇒ 8r – 3n – 3 =0………….2

We have 1 and 2 as

n – 4r ± 5 =0…………1

8r – 3n – 3 =0…………….2

Multiplying equation 1 by number 2

2n -8r +10 =0……………….3

2n -8r +10 =0

-3n – 8r – 3 =0

⇒ -n = -7

n =7 and r = 3

11. 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:

The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br

The general term for binomial (1+x)2n is

Tr+1 = 2nCr xr …………………..1

To find the coefficient of xn

r = n

Tn+1 = 2nCn xn

The coefficient of xn = 2nCn

The general term for binomial (1+x)2n-1 is

Tr+1 = 2n-1Cr xr

To find the coefficient of xn

Putting n = r

Tr+1 = 2n-1Cr xn

The coefficient of xn = 2n-1Cn

We have to prove

Coefficient of xn in (1+x)2n = 2 coefficient of xn in (1+x)2n-1

Consider LHS = 2nCn

12. Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6.

Solution:

The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br

Here a = 1, b = x and n = m

Putting the value

Tr+1 = m Cr 1m-r xr

m Cr xr

We need coefficient of x2

∴ putting r = 2

T2+1 = mC2 x2

The coefficient of x2 = mC2

Given that coefficient of x2 = mC2 = 6

⇒ m (m – 1) = 12

⇒ m2– m – 12 =0

⇒ m2– 4m + 3m – 12 =0

⇒ m (m – 4) + 3 (m – 4) = 0

⇒ (m+3) (m – 4) = 0

⇒ m = – 3, 4

We need positive value of m so m = 4

Miscellaneous Exercise Page No: 175

1. 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:

We know that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by

Tr+1 = nCr an-t br

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

T1 = nC0 an-0 b0 = an = 729….. 1

T2 = nC1 an-1 b1 = nan-1 b = 7290…. 2

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

Dividing 2 by 1 we get

$\frac{na^{n-1}b}{a^{n}} = \frac{7290}{729} \\ \frac{nb}{a} = 10$

Dividing 3 by 2 we get

$\frac{n(n-1)a^{n-2}b^{2}}{2na^{n-1}b} = \frac{30375}{7290} \\ \frac{(n-1)b}{2a} = \frac{30375}{7290} \\ \frac{(n-1)b}{a} = \frac{30375}{7290} \times 2 =\frac{25}{3} \\ \frac{(nb)}{a} -\frac{b}{a} = \frac{25}{3} \\ 10 -\frac{b}{a} = \frac{25}{3} \\ \frac{b}{a} = 10 – \frac{25}{3} = \frac{5}{3}$

From 4 and 5 we have

n. 5/3 = 10

n = 6

Substituting n = 6 in 1 we get

a6 = 729

a = 3

From 5 we have, b/3 = 5/3

b = 5

Thus a = 3, b = 5 and n = 76

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

Solution:

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

Solution:

(1 + 2x)6 = 6C0 + 6C1 (2x) + 6C2 (2x)2 + 6C3 (2x)3 + 6C4 (2x)4 + 6C5 (2x)5 + 6C6 (2x)6

= 1 + 6 (2x) + 15 (2x)2 + 20 (2x)3 + 15 (2x)4 + 6 (2x)5 + (2x)6

= 1 + 12 x + 60x2 + 160 x3 + 240 x4 + 192 x5 + 64x6

(1 – x)7 = 7C07C1 (x) + 7C2 (x)27C3 (x)3 + 7C4 (x)47C5 (x)5 + 7C6 (x)6 7C7 (x)7

= 1 – 7x + 21x2 – 35x3 + 35x4 – 21x5 + 7x6 – x7

(1 + 2x)6 (1 – x)7 = (1 + 12 x + 60x2 + 160 x3 + 240 x4 + 192 x5 + 64x6) (1 – 7x + 21x2 – 35x3 + 35x4 – 21x5 + 7x6 – x7)

192 – 21 = 171

Thus, the coefficient of x5 in the expression (1+2x)6(1-x)7 is 171.

4. 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:

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.

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 bn] is a natural number

This shows that (a – b) is a factor of (an – bn), where n is positive integer.

5. Evaluate

Solution:

Using binomial theorem the expression (a + b)6 and (a – b)6, can be expanded

(a + b)6 = 6C0 a6 + 6C1 a5 b + 6C2 a4 b2 + 6C3 a3 b3 + 6C4 a2 b4 + 6C5 a b5 + 6C6 b6

(a – b)6 = 6C0 a66C1 a5 b + 6C2 a4 b26C3 a3 b3 + 6C4 a2 b46C5 a b5 + 6C6 b6

Now (a + b)6 – (a – b)6 =6C0 a6 + 6C1 a5 b + 6C2 a4 b2 + 6C3 a3 b3 + 6C4 a2 b4 + 6C5 a b5 + 6C6 b6 – [6C0 a66C1 a5 b + 6C2 a4 b26C3 a3 b3 + 6C4 a2 b46C5 a b5 + 6C6 b6]

Now by substituting a = √3 and b = √2 we get

(√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 (√6) (198)

= 396 √6

6. Find the value of

Solution:

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

Solution:

0.99 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)55C1 (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

8. Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of  is √6: 1

Solution:

9. Expand using Binomial Theorem

Solution:

Using binomial theorem the given expression can be expanded as

Again by using binomial theorem to expand the above terms we get

From equation 1, 2 and 3 we get

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

Solution:

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

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

[3x2 + (-a (2x-3a))]3

= (3x2)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

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

Thus, (3x2 – 2ax + 3a2)3

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

 Also Access NCERT Exemplar for Class 11 Maths Chapter 8 CBSE Notes for Class 11 Maths Chapter 8

## NCERT Solutions for Class 11 Maths Chapter 8- Binomial Theorem

The Chapter 8 Binomial Theorem of NCERT Solutions for Class 11 covers the topics given below.

8.1 Introduction to Binomial Theorem

8.2 Binomial Theorem for Positive Integral Indices

Pascal’s Triangle

8.2.1 Binomial theorem for any positive integer n,

8.2.2 Some special cases

8.3 General and Middle Terms

Exercise 8.1 Solutions 14 Questions

Exercise 8.2 Solutions 12 Questions

Miscellaneous Exercise On Chapter 8 Solutions 10 Questions

## NCERT Solutions for Class 11 Maths Chapter 8- Binomial Theorem

The unit Algebra houses the chapter Binomial Theorem, adding up to 30 marks of the total 80 marks. A total of 3 exercises including the miscellaneous exercise is present in this chapter. Chapter 8 of NCERT Solutions for Class 11 Maths discusses the concepts provided underneath:

1. The expansion of a binomial for any positive integral n is given by the Binomial Theorem, which is (a+b)n = nC0 an + nC1 an – 1b + nC2 an – 2b2 + …+ nCn – 1a.bn – 1 + nCn bn .
2. The coefficients of the expansions are arranged in an array. This array is called Pascal’s triangle.
3. The general term of an expansion (a + b)n is Tr + 1 = nCr an – r . br

Therefore, it is thus assured that a student is thorough with the eighth chapter of Class 11, the Binomial Theorem, will be well versed in the history of Binomial Theorem, statement and proof of the binomial theorem for positive integral indices, Pascal’s triangle, General and middle term in binomial expansion as well as simple applications of Binomial theorem.