NCERT Solutions for Class 11 Science Maths Chapter 8 Binomial Theorem are provided here with simple step-by-step explanations. These solutions for Binomial Theorem are extremely popular among Class 11 Science students for Maths 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 Science Maths Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s NCERT Solutions. All NCERT Solutions for class Class 11 Science Maths are prepared by experts and are 100% accurate.

#### Page No 166:

#### Question 1:

Expand
the expression (1– 2*x*)^{5}

#### Answer:

By
using Binomial Theorem, the expression (1–
2*x*)^{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 (2*x* – 3)^{6 }

#### Answer:

By
using Binomial Theorem, the expression (2*x*
– 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}
– (*a*
– *b*)^{4}.
Hence, evaluate.

#### Answer:

Using
Binomial Theorem, the expressions, (*a*
+ *b*)^{4}
and (*a* –
*b*)^{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 *x*^{5}
in (*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 *x*^{5}
occurs in the (*r*
+ 1)^{th}
term of the expansion (*x*
+ 3)^{8},
we obtain

Comparing
the indices of *x*
in *x*^{5}
and in *T*_{r}_{
+1}, we obtain

*r*
= 3

Thus,
the coefficient of *x*^{5}
is

#### Page No 171:

#### Question 2:

Find
the coefficient of* a*^{5}*b*^{7}
in (*a* –
2*b*)^{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 *a*^{5}*b*^{7}
occurs in the (*r*
+ 1)^{th}
term of the expansion (*a*
– 2*b*)^{12},
we obtain

Comparing
the indices of *a*
and *b* in
*a*^{5}
b^{7 }and
in *T*_{r}_{
+1}, we obtain

*r*
= 7

Thus,
the coefficient of *a*^{5}*b*^{7}
is

#### Page No 171:

#### Question 3:

Write
the general term in the expansion of (*x*^{2}
– *y*)^{6}

#### 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 (*x*^{2}
– *y*^{6})
is

#### Page No 171:

#### Question 4:

Write
the general term in the expansion of (*x*^{2}
– *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(*x*^{2}
– *yx*)^{12}
is

#### Page No 171:

#### Question 5:

Find
the 4^{th}
term in the expansion of (*x*
– 2*y*)^{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 4^{th}
term in the expansion of (*x*
– 2*y*)^{12}
is

#### Page No 171:

#### Question 6:

Find
the 13^{th}
term in the expansion of.

#### Answer:

*r
*+ 1)^{th}
term, (*T*_{r}_{+1}),
in the binomial expansion of (*a *+
*b*)^{n}
is given by
.

Thus,
13^{th}
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 *x*^{5}*y*^{5}.

#### Page No 171:

#### Question 9:

In the expansion of (1 + *a*)^{m
+ n}, prove that coefficients of
*a*^{m}
and *a*^{n}
are equal.

#### Answer:

*r
*+ 1)^{th}
term, (*T*_{r}_{+1}),
in the binomial expansion of (*a *+
*b*)^{n}
is given by
.

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

Comparing
the indices of *a*
in *a*^{m}
and in *T*_{r
}_{+ 1},
we obtain

*r*
= *m*

Therefore,
the coefficient of *a*^{m}
is

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

Comparing
the indices of *a*
in *a*^{n}
and in *T*_{k}_{
+ 1}, we obtain

*k*
= *n*

Therefore,
the coefficient of *a*^{n}
is

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

#### Page No 171:

#### Question 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*.

#### 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

4*r
*– 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 *x*^{n}
in the expansion of (1 + *x*)^{2}^{n}
is twice the coefficient of *x*^{n}
in the expansion of (1 + *x*)^{2}^{n}^{–1
}.

#### Answer:

*r
*+ 1)^{th}
term, (*T*_{r}_{+1}),
in the binomial expansion of (*a *+
*b*)^{n}
is given by
.

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

Comparing
the indices of *x*
in *x*^{n}
and in *T*_{r}_{
+ 1}, we obtain

*r*
=* n*

Therefore,
the coefficient of* x*^{n}
in the expansion of (1 + *x*)^{2}^{n}
is

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

Comparing
the indices of *x*
in *x*^{n}
and *T*_{k}_{
+ 1}, we obtain

*k*
=* n*

Therefore,
the coefficient of* x*^{n}
in the expansion of (1 + *x*)^{2}^{n
}^{–1}
is

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

Therefore,
the coefficient of *x*^{n}
in the expansion of (1 + *x*)^{2}^{n}
is twice the coefficient of *x*^{n}
in the expansion of (1 + *x*)^{2}^{n}^{–1}.

Hence, proved.

#### Page No 171:

#### Question 12:

Find a positive
value of *m*
for which the coefficient of *x*^{2}
in the expansion

(1
+ *x*)^{m}
is 6.

#### Answer:

*r
*+ 1)^{th}
term, (*T*_{r}_{+1}),
in the binomial expansion of (*a *+
*b*)^{n}
is given by
.

Assuming
that *x*^{2}
occurs in the (*r *+
1)^{th}
term of the expansion (1 +*x*)^{m},
we obtain

Comparing
the indices of *x*
in *x*^{2}
and in *T*_{r}_{
+ 1}, we obtain

*r*
= 2

Therefore,
the coefficient of *x*^{2}
is.

It
is given that the coefficient of *x*^{2}
in the expansion (1 + *x*)^{m}
is 6.

Thus,
the positive value of *m*,
for which the coefficient of *x*^{2}
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:

*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

*a*^{6}
= 729

From (5), we obtain

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

#### Page No 175:

#### Question 2:

Find *a*
if the coefficients of *x*^{2}
and *x*^{3}
in the expansion of (3 + *ax*)^{9}
are equal.

#### Answer:

*r
*+ 1)^{th}
term, (*T*_{r}_{+1}),
in the binomial expansion of (*a *+
*b*)^{n}
is given by
.

Assuming
that *x*^{2}
occurs in the (*r*
+ 1)^{th}
term in the expansion of (3 + *ax*)^{9},
we obtain

Comparing
the indices of *x*
in *x*^{2}
and in *T*_{r}_{
+ 1}, we obtain

*r*
= 2

Thus,
the coefficient of *x*^{2}
is

Assuming
that *x*^{3}
occurs in the (*k*
+ 1)^{th}
term in the expansion of (3 + *ax*)^{9},
we obtain

Comparing
the indices of *x*
in *x*^{3}
and in *T*_{k}_{+
1}, we obtain

*k
*= 3

Thus,
the coefficient of *x*^{3}
is

It
is given that the coefficients of *x*^{2}
and *x*^{3}
are the same.

Thus,
the required value of *a*
is.

#### Page No 175:

#### Question 3:

Find the coefficient of *x*^{5}
in the product (1 + 2*x*)^{6}
(1 – *x*)^{7}
using binomial theorem.

#### Answer:

Using
Binomial Theorem, the expressions, (1 + 2*x*)^{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 *x*^{5},
are required.

The
terms containing *x*^{5}
are

Thus,
the coefficient of *x*^{5}
in the given product is 171.

#### Page No 175:

#### Question 4:

If* a* and *b* are distinct integers, prove that *a* – *b* is a factor of *a*^{n} – *b*^{n}, whenever *n* is a positive integer.

[**Hint:** write *a*^{n} = (*a* –* b *+ *b*)^{n} and expand]

#### Answer:

In
order to prove that (*a*
– *b*)
is a factor of (*a*^{n}
– *b*^{n}),
it has to be proved that

*a*^{n}
– *b*^{n}
= *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 (*a*^{n}
– *b*^{n}),
where *n* is
a positive integer.

#### Page No 175:

#### Question 5:

Evaluate.

#### Answer:

Firstly,
the expression (*a*
+ *b*)^{6}
– (*a*
– *b*)^{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}
+ (*x* –
*y*)^{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 11