Question

Find the sum of the series $1+2^{2}x+3^2{x}^2+4^2{x}^3+...\infty \left|x\right|<1.$

• 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

Answer: (B)

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

Let $S_{\infty }=1+4x+9x^2+16x^3+...\infty$ $...\left(1\right)$
Now multiplying by $x$ throughout in $\left(1\right)$; we get
$x{S}_{\infty }=1+4x^2+9x^3+16x^4+...\infty$ $...\left(2\right)$
Subtracting equation $\left(2\right)$ from $\left(1\right)$; we get
$(1-x{)S}_{\infty }=1+4x+9x^2+16x^3+....\infty -x-4x^2-9x^3-{16x}^4-...\infty$
$(1-x)S_{\infty }=1+3x+5x^2+7x^3+....\infty$ $...\left(3\right)$
Again, multiplying by $x$ throughout in $\left(3\right)$; we get
$x(1-x)S_{\infty }=x+3x^2+5x^3+7x^4+....\infty$ $...\left(4\right)$
Subtracting equation $\left(4\right)$ from $\left(3\right)$; we get
$(1-x)S_{\infty }-x(1-x)S_{\infty }=1+x+2x+2x^2+2x^3+....\infty$
Notice that the series $2x+2x^2+2x^3+...\infty$ is
geometric series with the first term $a=2x$ and the common ratio
$r=x$.
Now, use the formula for the sum of an infinite geometric series.
$(1-x)S_{\infty }-x(1-x)S_{\infty }=1+\frac{2x}{(1-x)}$, for$\left|x\right|<1$
$\Rightarrow (1-x)(1-x)S_{\infty }=\frac{1+x}{(1-x)}$, for$\left|x\right|<1$
$\Rightarrow S_{\infty }=\frac{1+x}{{(1-x)}^3}$, for$\left|x\right|<1$