Question

If $\left|x\right|<1$, then the sum of the series $1+2x+3x^2+4x^3+.......\infty$ will be

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

Answer 1

The correct option is D

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

Explanation for the correct answer.

Find the sum of the series.

Let, sum of the series is $S$:

$S=1+2x+3x^2+4x^3+.......\infty ...\left(1\right)$

Now, multiply $x$ in both side

$Sx=x+2x^2+3x^3+4x^4+.....\infty ...\left(2\right)$

On subtracting $\left(2\right)$ by $\left(1\right)$.

$\begin{array}{rcl}S-Sx& =& 1-x+2x-2x^2+3x^2-3x^3+.....\infty \\ S\left(1-x\right)& =& 1+x+x^2+x^3+....\infty \\ S\left(1-x\right)& =& \frac{1}{\left(1-x\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left[\because sumofinfiniteGPis\frac{a}{\left(1-r\right)}\right]\\ S& =& \frac{1}{{\left(1-x\right)}^2}\end{array}$

Hence, the correct option is D.