Question

The coefficient of $x^5$ in $(1+2x+3x^2+.....{)}^{-3/2}$ is

• A. $21$
• B. $25$
• C. $26$
• D. $Noneofthese$

Answer 1

Answer: (C), (D)

$Noneofthese$
$26$

Solution:- (D) None of these

Let $S=1+2x+3x^2+.....$

$xS=x+2x^2+3x^3+.......$

$\therefore S-xS=1+x+x^2+x^3+.......$

As the above sequence is a G.P. of infinite terms.

$\therefore S(1-x)=\frac{1}{1-x}[\because \text{Sum of infinite terms of g.p. of common ration}r=\frac{1}{1-r}]$

$\Rightarrow S=\frac{1}{{(1-x)}^2}={(1-x)}^{-2}$

$\therefore {(1+2x+3x^2+.....)}^{-\frac{3}{2}}={\left({(1-x)}^{-2}\right)}^{-\frac{3}{2}}={(1-x)}^3$

As ${(1-x)}^3=1-x^3-3x+3x^2$

Since there is no term of $x^5$ in the above expansion.

Hence the coefficient of the $x^5$ in the above expression is $0$.