NCERT Solutions for Class 11 Humanities Math Chapter 8 Binomial Theorem are provided here with simple step-by-step explanations. These solutions for Binomial Theorem are extremely popular among Class 11 Humanities students for Math Binomial Theorem Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 11 Humanities Math Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class Class 11 Humanities Math are prepared by experts and are 100% accurate.

Page No 166:

Question 1:

Expand the expression (1– 2x)5

Answer:

By using Binomial Theorem, the expression (1– 2x)5 can be expanded as

Page No 166:

Question 2:

Expand the expression

Answer:

By using Binomial Theorem, the expression can be expanded as

Page No 166:

Question 3:

Expand the expression (2x – 3)6

Answer:

By using Binomial Theorem, the expression (2x – 3)6 can be expanded as



Page No 167:

Question 4:

Expand the expression

Answer:

By using Binomial Theorem, the expression can be expanded as

Page No 167:

Question 5:

Expand

Answer:

By using Binomial Theorem, the expression can be expanded as

Page No 167:

Question 6:

Using Binomial Theorem, evaluate (96)3

Answer:

96 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, binomial theorem can be applied.

It can be written that, 96 = 100 – 4

Page No 167:

Question 7:

Using Binomial Theorem, evaluate (102)5

Answer:

102 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.

It can be written that, 102 = 100 + 2

Page No 167:

Question 8:

Using Binomial Theorem, evaluate (101)4

Answer:

101 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.

It can be written that, 101 = 100 + 1

Page No 167:

Question 9:

Using Binomial Theorem, evaluate (99)5

Answer:

99 can be written as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.

It can be written that, 99 = 100 – 1

Page No 167:

Question 10:

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

Answer:

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

Page No 167:

Question 11:

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

Answer:

Using Binomial Theorem, the expressions, (a + b)4 and (ab)4, can be expanded as

Page No 167:

Question 12:

Find (x + 1)6 + (x – 1)6. Hence or otherwise evaluate.

Answer:

Using Binomial Theorem, the expressions, (x + 1)6 and (x – 1)6, can be expanded as

By putting, we obtain

Page No 167:

Question 13:

Show that is divisible by 64, whenever n is a positive integer.

Answer:

In order to show that is divisible by 64, it has to be proved that,

, where k is some natural number

By Binomial Theorem,

For a = 8 and m = n + 1, we obtain

Thus, is divisible by 64, whenever n is a positive integer.

Page No 167:

Question 14:

Prove that.

Answer:

By Binomial Theorem,

By putting b = 3 and a = 1 in the above equation, we obtain

Hence, proved.



Page No 171:

Question 1:

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

Answer:

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

Assuming that x5 occurs in the (r + 1)th term of the expansion (x + 3)8, we obtain

Comparing the indices of x in x5 and in Tr +1, we obtain

r = 3

Thus, the coefficient of x5 is

Page No 171:

Question 2:

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

Answer:

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

Assuming that a5b7 occurs in the (r + 1)th term of the expansion (a – 2b)12, we obtain

Comparing the indices of a and b in a5 b7 and in Tr +1, we obtain

r = 7

Thus, the coefficient of a5b7 is

Page No 171:

Question 3:

Write the general term in the expansion of (x2y)6

Answer:

It is known that the general term Tr+1 {which is the (r + 1)th term} in the binomial expansion of (a + b)n is given by .

Thus, the general term in the expansion of (x2y6) is

Page No 171:

Question 4:

Write the general term in the expansion of (x2yx)12, x 0

Answer:

It is known that the general term Tr+1 {which is the (r + 1)th term} in the binomial expansion of (a + b)n is given by .

Thus, the general term in the expansion of(x2yx)12 is

Page No 171:

Question 5:

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

Answer:

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

Thus, the 4th term in the expansion of (x – 2y)12 is

Page No 171:

Question 6:

Find the 13th term in the expansion of.

Answer:

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

Thus, 13th term in the expansion of is

Page No 171:

Question 7:

Find the middle terms in the expansions of

Answer:

It is known that in the expansion of (a + b)n, if n is odd, then there are two middle terms, namely, term and term.

Therefore, the middle terms in the expansion of are term and term

Thus, the middle terms in the expansion of are .

Page No 171:

Question 8:

Find the middle terms in the expansions of

Answer:

It is known that in the expansion (a + b)n, if n is even, then the middle term is term.

Therefore, the middle term in the expansion of is term

Thus, the middle term in the expansion of is 61236 x5y5.

Page No 171:

Question 9:

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

Answer:

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

Assuming that am occurs in the (r + 1)th term of the expansion (1 + a)m + n, we obtain

Comparing the indices of a in am and in Tr + 1, we obtain

r = m

Therefore, the coefficient of am is

Assuming that an occurs in the (k + 1)th term of the expansion (1 + a)m+n, we obtain

Comparing the indices of a in an and in Tk + 1, we obtain

k = n

Therefore, the coefficient of an is

Thus, from (1) and (2), it can be observed that the coefficients of am and an in the expansion of (1 + a)m + n are equal.

Page No 171:

Question 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.

Answer:

It is known that (k + 1)th term, (Tk+1), in the binomial expansion of (a + b)n is given by .

Therefore, (r – 1)th term in the expansion of (x + 1)n is

r th term in the expansion of (x + 1)n is

(r + 1)th term in the expansion of (x + 1)n is

Therefore, the coefficients of the (r – 1)th, rth, and (r + 1)th terms in the expansion of (x + 1)n are respectively. Since these coefficients are in the ratio 1:3:5, we obtain

Multiplying (1) by 3 and subtracting it from (2), we obtain

4r – 12 = 0

r = 3

Putting the value of r in (1), we obtain

n – 12 + 5 = 0

n = 7

Thus, n = 7 and r = 3

Page No 171:

Question 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 .

Answer:

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

Assuming that xn occurs in the (r + 1)th term of the expansion of (1 + x)2n, we obtain

Comparing the indices of x in xn and in Tr + 1, we obtain

r = n

Therefore, the coefficient of xn in the expansion of (1 + x)2n is

Assuming that xn occurs in the (k +1)th term of the expansion (1 + x)2n – 1, we obtain

Comparing the indices of x in xn and Tk + 1, we obtain

k = n

Therefore, the coefficient of xn in the expansion of (1 + x)2n –1 is

From (1) and (2), it is observed that

Therefore, 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.

Hence, proved.

Page No 171:

Question 12:

Find a positive value of m for which the coefficient of x2 in the expansion

(1 + x)m is 6.

Answer:

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

Assuming that x2 occurs in the (r + 1)th term of the expansion (1 +x)m, we obtain

Comparing the indices of x in x2 and in Tr + 1, we obtain

r = 2

Therefore, the coefficient of x2 is.

It is given that the coefficient of x2 in the expansion (1 + x)m is 6.

Thus, the positive value of m, for which the coefficient of x2 in the expansion

(1 + x)m is 6, is 4.



Page No 175:

Question 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.

Answer:

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

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

Therefore, we obtain

Dividing (2) by (1), we obtain

Dividing (3) by (2), we obtain

From (4) and (5), we obtain

Substituting n = 6 in equation (1), we obtain

a6 = 729

From (5), we obtain

Thus, a = 3, b = 5, and n = 6.

Page No 175:

Question 2:

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

Answer:

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

Assuming that x2 occurs in the (r + 1)th term in the expansion of (3 + ax)9, we obtain

Comparing the indices of x in x2 and in Tr + 1, we obtain

r = 2

Thus, the coefficient of x2 is

Assuming that x3 occurs in the (k + 1)th term in the expansion of (3 + ax)9, we obtain

Comparing the indices of x in x3 and in Tk+ 1, we obtain

k = 3

Thus, the coefficient of x3 is

It is given that the coefficients of x2 and x3 are the same.

Thus, the required value of a is.

Page No 175:

Question 3:

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

Answer:

Using Binomial Theorem, the expressions, (1 + 2x)6 and (1 – x)7, can be expanded as

The complete multiplication of the two brackets is not required to be carried out. Only those terms, which involve x5, are required.

The terms containing x5 are

Thus, the coefficient of x5 in the given product is 171.

Page No 175:

Question 4:

If a and b are distinct integers, prove that ab is a factor of anbn, whenever n is a positive integer.

[Hint: write an = (a b + b)n and expand]

Answer:

In order to prove that (ab) is a factor of (anbn), it has to be proved that

anbn = k (ab), where k is some natural number

It can be written that, a = ab + b

This shows that (ab) is a factor of (anbn), where n is a positive integer.

Page No 175:

Question 5:

Evaluate.

Answer:

Firstly, the expression (a + b)6 – (ab)6 is simplified by using Binomial Theorem.

This can be done as

Page No 175:

Question 6:

Find the value of.

Answer:

Firstly, the expression (x + y)4 + (xy)4 is simplified by using Binomial Theorem.

This can be done as

Page No 175:

Question 7:

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

Answer:

0.99 = 1 – 0.01

Thus, the value of (0.99)5 is approximately 0.951.

Page No 175:

Question 8:

Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of

Answer:

In the expansion, ,

Fifth term from the beginning

Fifth term from the end

Therefore, it is evident that in the expansion of, the fifth term from the beginning is and the fifth term from the end is.

It is given that the ratio of the fifth term from the beginning to the fifth term from the end is. Therefore, from (1) and (2), we obtain

Thus, the value of n is 10.



Page No 176:

Question 9:

Expand using Binomial Theorem.

Answer:

Using Binomial Theorem, the given expression can be expanded as

Again by using Binomial Theorem, we obtain

From (1), (2), and (3), we obtain

Page No 176:

Question 10:

Find the expansion of using binomial theorem.

Answer:

Using Binomial Theorem, the given expression can be expanded as

Again by using Binomial Theorem, we obtain

From (1) and (2), we obtain



View NCERT Solutions for all chapters of Class 14