Question

Find the sum of the series $1+2x+3x^2+4x^3+...$

• A. $\frac{-1}{{(1-x)}^2}$
• B. $\frac{1}{{(1-x)}^2}$
• C. $\frac{-1}{{(1+x)}^2}$
• D. $\frac{1}{{(1+x)}^2}$

Answer 1

Answer: (B)

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

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