NCERT Solution for Class11 Maths Chapter 8 Binomial theorem

Exercise 8.1 : Solutions of Questions on Page Number : 166

NCERT Solution for Class11 Maths Chapter 8 Binomial theorem

Q1 : Expand the expression (1- 2x)⁵
Answer :
By using Binomial Theorem, the expression (1- 2x)5 can be expanded as

Q2 : Expand the expression
Answer :
By using Binomial Theorem, the expression can be expanded as

Q3 : Expand the expression (2x – 3)⁶
Answer :
By using Binomial Theorem, the expression (2x- 3)⁶ can be expanded as

Q4 : Expand the expression
Answer :
By using Binomial Theorem, the expression can be expanded

Q5 : Expand
Answer :
By using Binomial Theorem, the expression can be expanded as

Q6 : 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

Q7 : Using Binomial Theorem, evaluate (102)⁵
Answer :
102can 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

Q8 : 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

Q9 : 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

Q10 : 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)10000can be obtained as

Q11 : Find (a + b)⁴– (a- b)⁴. Hence, evaluate
Answer :
Using Binomial Theorem, the expressions, (a+ b)4and (a – b)4, can be expanded a

Q12 : 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

Q13 : Show that is divisible by 64, whenever nis 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.

Q14 : Prove that
Answer :
By Binomial Theorem,

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

Hence, proved.

Exercise 8.2 : Solutions of Questions on Page Number : 171
Q1 : Find the coefficient of x5in (x + 3)⁸
Answer :
It is known that (r + 1)th term, (T_r+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 x⁵and in Tr+1, we obtain
r= 3
Thus, the coefficient of x⁵ is

Q2 : Find the coefficient of a⁵b⁷in (a – 2b)12
Answer :
It is known that (r + 1)th term, (T_r+1), in the binomial expansion of (a + b)ⁿ is given by
Assuming that a⁵b⁷ occurs in the (r+ 1)thterm of the expansion (a- 2b)¹², we obtain

Comparing the indices of aand b in a⁵b⁷ and in Tr+1, we obtain
r= 7
Thus, the coefficient of a⁵b⁷ is

Q3 : Write the general term in the expansion of (x²– y)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 (x²– y⁶) is

Q4 : Write the general term in the expansion of (x²– yx)¹², x ≠0
Answer :
It is known that the general term T_r+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(x²-yx)¹² is

Q5 : Find the 4th term in the expansion of (x- 2y)¹² .
Answer :
It is known that (r + 1)th term, (T_r+1), in the binomial expansion of (a + b)ⁿ is given by .
Thus, the 4thterm in the expansion of (x- 2y)¹² is

Q6 : Find the 13th term in the expansion of .
Answer :
It is known that (r + 1)th term, (T_r+1), in the binomial expansion of (a + b)ⁿ is given by .
Thus, 13th term in the expansion of is

Q7 : 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 .

Q8 : 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.

Q9 : In the expansion of (1 + a)^{m + n,} prove that coefficients of a^m and aⁿ are equal.
Answer :
It is known that (r + 1)th term, (T_r+1), in the binomial expansion of (a + b)ⁿ is given by .
Assuming that amoccurs in the (r+ 1)th term of the expansion (1 + a)^{m+ n,} we obtain

Comparing the indices of ain amand in T_{r + 1}, we obtain
r= m
Therefore, the coefficient of a^m is

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

Comparing the indices of ain anand in T_{k+ 1}, we obtain
k= n
Therefore, the coefficient of aⁿ is

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

Q10 : 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 nand r.
Answer :
It is known that (k + 1)th term, (T_k+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, r^th, 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 rin (1), we obtain
n- 12 + 5 = 0
⇒ n= 7
Thus, n = 7 and r = 3

Q11 : 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, (T_r+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 T_{r+ 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 T_{k+ 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.

Q12 : 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, (T_r+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 T_{r+ 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.

Exercise Miscellaneous : Solutions of Questions on Page Number : 175
Q1 : Find a, band 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, (T_r+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.

Q2 : Find a if the coefficients of x² and x³ in the expansion of (3 + ax)⁹are equal.
Answer :
It is known that (r + 1)^th term, (T_r+1), in the binomial expansion of (a + b)_n is given by .
Assuming that x2occurs in the (r+ 1)thterm in the expansion of (3 + ax)⁹, we obtain

Comparing the indices of x in x² and in T_{r+ 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 T_{k+ 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.

Q3 : Find the coefficient of x5in the product (1 + 2x)⁶(1 – x)⁷using 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.

Q4 : 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]
Answer :
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
It can be written that, a= a – b + b

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

Q5 : Evaluate .
Answer :
Firstly, the expression (a+ b)⁶– (a- b)⁶ is simplified by using Binomial Theorem.
This can be done as

Q6 : Find the value of .
Answer :
Firstly, the expression (x+ y)⁴+ (x – y)⁴is simplified by using Binomial Theorem.
This can be done as

Q7 : 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.

Q8 : 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.

Q9 : 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

Q10 : 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