NCERT Solutions for Class 11 Maths Chapter 8 - Binomial Theorem Exercise 8.2

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 7.

The NCERT Solutions are compiled and developed as per the latest CBSE syllabus. The NCERT Solutions contain detailed step-by-step explanations for all the problems that are present in the textbook. Exercise 8.2 of NCERT Solutions for Class 11 Maths Chapter 8 – Binomial Theorem is based on the topic, General and Middle Terms. In the binomial expansion of (a + b)n, we observe that the first term is nC0an, the second term is nC1an–1b, the third term is nC2an–2b2, and so on. Looking at the pattern of the successive terms, we can say that the (r + 1)th term is nCran–rbr. The (r + 1)th term is also called the general term of the expansion (a + b)n. It is denoted by Tr+1. Thus Tr+1 = nCran–rbr.

The NCERT solutions are prepared with the utmost care by the subject-matter experts at BYJU’S for the students to practise and score well in the annual exams. Students can view as well as download the NCERT Solutions for Class 11 Maths Chapter 8 in the link below.

NCERT Solutions for Class 11 Maths Chapter 8 – Binomial Theorem Exercise 8.2

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Access NCERT Solutions for Class 11 Maths Chapter 8 – Exercise 8.2

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)

To find out x5

We have to equate x5= x8-r

⇒ r= 3

Putting the value of r in (i), we get

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 12

= 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

Equate a12-r =a5

r = 7

Putting r = 7 in (i)

T8 = 12C7 a5 (-2b)7

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 13

= -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

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 14

= -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

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 15

= -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

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 16

= -1760 x9 y3

6. Find the 13th term in the expansion of 

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 17

Solution:

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 18

Find the middle terms in the expansions of
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 19

Solution:

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 20

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 21

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 22

Solution:

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 23

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 24

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, the general term for the expression is,

Tr+1 =  m+n Cr ar

Now, for the coefficient of am

Tm+1 =  m+n Cm am

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

So, the coefficient is m+n C m

Similarly, the coefficient of an is m+n C n

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 25

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,

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 26

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 27

By cross multiplication,

⇒ 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

Adding equations 2 and 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

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 29

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 the coefficient of x2

∴ putting r = 2

T2+1 = mC2 x2

The coefficient of x2 = mC2

Given that coefficient of x2 = mC2 = 6

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⇒ 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 the positive value of m, so m = 4


Access Other Exercise Solutions of Class 11 Maths Chapter 8 – Binomial Theorem

Exercise 8.1 Solutions 14 Questions

Miscellaneous Exercise on Chapter 8 Solutions 10 Questions

Also explore – NCERT Class 11 Maths Solutions

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