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NCERT · Class 11 · Maths

Chapter 8: Binomial Theorem Exercise 8.2

Practice each question, then open the answer or worked solution.

Questions

12 questions
Q1Written answer

x5 in (x + 3)8

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Worked 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

Q2Written answer

a5b7 in (a – 2b)12.

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Worked 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

Q3Written answer

(x2 – y)6

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Worked 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

Q4Written answer

(x2 – y x)12, x ≠ 0.

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Worked 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

Q5Written answer

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

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Worked 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

Q6Written answer

Find the 13th term in the expansion of 

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

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Worked solution

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

Q7Written answer

Find the middle terms in the expansions of

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

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Worked 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

Q8Written answer

Find the middle terms in the expansions of

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

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Worked 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

Q9Written answer

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

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Worked 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

Q10Written answer

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.

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Worked 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