Question

The sum of the series $1+2^{2}x+3^2{x}^2+4^2{x}^3+...\infty$, where $|x|<1$, is

• A. $\frac{(1-x)}{(1-x{)}^3}$
• B. $\frac{(1+x)}{(1-x{)}^3}$
• C. $\frac{(1+x)}{(1+x{)}^3}$
• D. none of these

Answer 1

The correct option is B $\frac{(1+x)}{(1-x{)}^3}$

Let $S_{\infty }=1+2^{2}x+3^2{x}^2+4^2{x}^3+\dots \infty$
$S_{\infty }=1+4x+9x^2+16x^3+\dots \infty \to \left(1\right)$
$\therefore x{S}_{\infty }=x+4x^2+9x^3+\dots \infty \to \left(2\right)$
Subtracting $\left(2\right)$ from $\left(1\right)$ by shifting one place, we get
$(1-x)S_{\infty }=1+3x+5x^2+7x^3+\dots \infty \to \left(3\right)$
Again,
$x(1-x)S_{\infty }=x+3x^2+5x^3+\dots \infty \to \left(4\right)$
Subtracting $\left(4\right)$ from $\left(3\right)$, we get
$(1-x)(1-x)S_{\infty }=1+2x+2x^2+2x^3+\dots \infty$
$(1-x{)}^2{S}_{\infty }=1+\frac{2x}{1-x}$ $=\frac{1+x}{1-x}$
$\Rightarrow S_{\infty }=\frac{(1+x)}{(1-x{)}^3}$