# RBSE Solutions for Class 11 Maths Chapter 7 Binomial Theorem Ex 7.4

## Rajasthan Board RBSE Class 11 Maths Chapter 7 Binomial Theorem Ex 7.4

Question 1.
Expand the following Binomials upto fourth term:

Solution:
Expansion upto four terms of (1 + x2)-2

Question 2.
Find the required terms in the following expansions:
(i) Fourth term of (1 – 3x)-1/3
(ii) Seventh term of (1 + x)5/2
(iii) Eighth term of (1 + 2x)-1/2
Solution:
(i) Fourth term of (1 – 3x)-1/3
= Fourth term of (1 – 3x)-1/3
= Fourth term of {1 + (- 3x)}1/3

(ii) Seventh term of (1 + x)5/2

(iii) Eighth term of (1 + 2x)-1/2

Question 3.
Find the general term of the following expansions:
(i) (a3 – x3)2/3
(ii) (1 – 2x)-3/2
(iii) (1 – x)-p/q
Solution:
(i) General term in the expansion of (a3 – x3)2/3

(ii) General term in the expansion of (1 – 2x)-3/2

(iii) General term of in the expansion of (1 – x)-p/q

Question 4.
If x < 3, find the coefficient of (3 – x)-8 in the expansion of (3 – x)– 5.
Solution:
(x + 1)th term in the expansion of (3 – x)-8

Question 5.
Find the coefficient of x6 in the expansion of (a + 2bx2)-3
Solution :
(r + 1)th term in the expansion of (a + 2bx2)-3

For the coefficient of x6 in the this term
2r = 6 ⇒ r = 3
Hence, from equation (i), coefficient of x6 in the expansion of (a + 2bx2)-3

Question 6.
Find the coefficient of x10 in the expansion of

Solution:

Question 7.
Find the coefficient of xr in the expansion of (1 – 2x + 3x2 – 4x3 + …)n and if x = $$\frac { 1 }{ 2 }$$ and n = 1, then find the value of the expression.
Solution:
we know that (1 + x)-2 = 1 – 2x + 3x2 – 4x3 +…+ (-1)r (r+1)xr+ ….. (i)
Given progression (1 – 2x + 3x2 – 4x3 +…)n
(1 – 2x + 3x2 – 4x3 +…)n
= {(1 + x)-2}n  [From equation (i)]
= (1 + x)-2n

Question 8.
Prove that (1 + x + x2 + x3 + …)2 = 1 + 2x + 3x2 + …
Solution:
We know that
1 + x + x2 + x3 + … = (1 – x)-1  …. (i)
(1 + x + x2 + x3 + …)2
= {(1 – x)-1}2  [From equation (i)]
= (1 – x)2

Question 9.
Prove that (1 + x + x2 + x3 +….)  (1 + 3x + 6x2+ …)
= (1 + 2x + 3x2 + …)2
Solution:
We know that
(1 – x)-1 = 1 + x + x2 + x3 +…           … (i)
and (1 – x)-3 = 1 + 3x + 6x2 + 10x3 + … (ii)
L.H.S. = (1 + x + x2 + x3 +…)(1 + 3x + 6x2 + …)
= {(1 – x)-1} {(1 – x)-3}
= {(1 – x)2}-2
= {(1 – x)-2}2
= (1 + 2x + 3x2 +…)2
= R.H.S.
Hence Proved.

Question 10.
If x = 2y + 3y2 + 4y3 + …then express y in series of as ascending powers of x.
Solution:
⇒ x = 2y + 3y2 + 4y2 + …
⇒ x = (1 + 2y + 3y2 + 4y3 + …)- 1
[∵ (1 – x)-2 = 1 + 2x + 3x2 + …]
⇒ (x+ 1) = (1 -y)-2