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 ⁿC₁a^n–1b, the third term is ⁿC₂a^n–2b², and so on. Looking at the pattern of the successive terms, we can say that the (r + 1)^th term is ⁿC_ra^n–rb^r. The (r + 1)^th term is also called the general term of the expansion (a + b)ⁿ. It is denoted by T_r+1. Thus T_r+1 = ⁿC_ra^n–rb^r.

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 T_r+1 in the binomial expansion is given by T_r+1 = ⁿ C _r a^n-r b^r

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

T_r+1 = ⁸C_r x^8-r 3^r…………… (i)

To find out x⁵

We have to equate x⁵= x^8-r

⇒ r= 3

Putting the 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 T_r+1 in the binomial expansion is given by T_r+1 = ⁿ C _r a^n-r b^r

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

Substituting the values, we get

T_r+1 = ¹²C_r a^12-r (-2b)^r………. (i)

To find a⁵

Equate a^12-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 T_r+1 in the binomial expansion is given by

T_r+1 = ⁿ C _r a^n-r b^r…….. (i)

Here, a = x² , n = 6 and b = -y

Putting values in (i)

T_r+1 = ⁶C_r x ^2(6-r) (-1)^r y^r

= -1^{r 6}c_r .x^{12 – 2r}. y^r

4. (x² – y x)¹², x ≠ 0.

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿ C _r a^n-r b^r

Here, n = 12, a= x² and b = -y x

Substituting the values, we get

T_n+1 =¹²C_r × x^2(12-r) (-1)^r y^r x^r

= -1^{r 12}c_r .x^{24 –2r}. y^r

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

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿ C _r a^n-r b^r

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 13^th 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 a^m and aⁿ are equal.

Solution:

We know that the general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

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

Substituting the values in the general form,

T_r+1 = ^m+n C_r 1^m+n-r a^r

= ^m+n C_r a^r…………. (i)

Now, the general term for the expression is,

T_r+1 =  ^m+n C_r a^r

Now, for the coefficient of a^m

T_m+1 =  ^m+n C_m a^m

Hence, for the coefficient of a^m, the value of r = m

So, the coefficient is ^m+n C _m

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

10. 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:

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

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) = ⁿC_r 1^n-r x^r

T_(r+1) = ⁿC_r x^r

The coefficient of (r+1)^th term is ⁿC_r

The r^th term is given by (r-1)^th term

T_(r+1-1) = ⁿC_r-1 x^r-1

T_r = ⁿC_r-1 x^r-1

∴ the coefficient of r^th term is ⁿC_r-1

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

T_r-2+1 = ⁿC_r-2 x^r-2

T_r-1 = ⁿC_r-2 x^r-2

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

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

Therefore,

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 T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

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

T_r+1 = ²ⁿC_r x^r …………………..1

To find the coefficient of xⁿ

r = n

T_n+1 = ²ⁿC_n xⁿ

The coefficient of xⁿ = ²ⁿC_n

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

T_r+1 = ^2n-1C_r x^r

To find the coefficient of xⁿ

Putting n = r

T_r+1 = ^2n-1C_r xⁿ

The coefficient of xⁿ = ^2n-1C_n

We have to prove

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

Consider LHS = ²ⁿC_n

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 T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

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

Putting the value,

T_r+1 = ^m C_r 1^m-r x^r

= ^m C_r x^r

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

⇒ 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