Question

The sum of the infinite series $1^2+2^{2}x+3^2{x}^2+....$ is

• A. $(1+x)/{(1-x)}^3$
• B. $(1+x)/(1-x)$
• C. $x/{(1-x)}^3$
• D. $1/{(1-x)}^3$

Answer 1

The correct option is A $(1+x)/{(1-x)}^3$

$1^2+2^{2}x+3^2{x}^2$

Let $5=1+4x+9x^2.........\infty$ ---------(1)

$\therefore 5x=x+4x^2+9x^3.........\infty$ ---------(2)

Subtract $\left(1\right)$ and $\left(2\right)$

$5(1-x)=1+3x+5x^2+7x^3.........\infty$------------(3)

equation $\left(3\right)$ is $A.G.P$ with

$a=1,d=2,r=x$

$\therefore$ sum of $A.C.P$

sum $R=\frac{a}{1-x}+\frac{dr}{(1-r{)}^2}$

$\therefore$ subtracting values

$R=\frac{1}{1-x}+\frac{2x}{(1-x{)}^2}$

$\therefore R=\frac{1+x}{(1-x{)}^2}$

Now $R=S(1-x)$

$\therefore S=\frac{R}{1-x}$

$\therefore S=\frac{1+x}{(1-x{)}^3}$