Free download NCERT Solutions for Class 11 Maths Chapter 8** Binomial Theorem Ex 8.1, Ex 8.2, and Miscellaneous Exercise** PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

Topics and Sub Topics in Class 11 Maths Chapter 8 Binomial Theorem:

Section Name | Topic Name |

8 | Binomial Theorem |

8.1 | Introduction |

8.2 | Binomial Theorem for Positive Integral Indices |

8.3 | General and Middle Terms |

**Contents**show

**NCERT Solutions for Class 11 Maths Chapter 8** – Binomial Theorem

**Question 1:**

Expand the expression (1– 2*x*)^{5}

**Answer:**

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

**Question 2:**

Expand the expression

**Answer:**

By using Binomial Theorem, the expression can be expanded as

**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

**Question 5:**

Expand

**Answer:**

By using Binomial Theorem, the expression can be expanded as

**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

**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

**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

**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

**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

**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

**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

**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.

**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

**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

**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

**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

**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

**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

**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 .

**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}.

**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.

**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

**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.

**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.

**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.

**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.

**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.

**Question 5:**

Evaluate.

**Answer:**

Firstly, the expression (*a* + *b*)^{6} – (*a* – *b*)^{6} is simplified by using Binomial Theorem.

This can be done as

**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

**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.

**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

**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

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