Question

The sum of the series $1+3x+6x^2+10x^3+.....\infty$ will be

• A. $\frac{1}{{\left(1-x\right)}^2}$
• B. $\frac{1}{\left(1-x\right)}$
• C. $\frac{1}{{\left(1+x\right)}^2}$
• D. $\frac{1}{{\left(1-x\right)}^3}$

Answer 1

The correct option is D

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

Explanation for the correct option

Let $S=1+3x+6x^2+10x^3+.....-------\left(1\right)$

multiply $S$by $x$

$xS=x+3x^2+6x^3+10x^4+........-------\left(2\right)$

subtracting $\left(2\right)$ from $\left(1\right)$

$$\begin{gathered} S-xS=1+3x+6x^2+10x^3+.....-\left(x+3x^2+6x^3+10x^4+........\right) \\ \Rightarrow S(1-x)=1+2x+3x^2+4x^3+..........--------\left(3\right) \end{gathered}$$

multiply $\left(3\right)$ by $x$

$xS(1-x)=x+2x^2+3x^3+4x^4+..........--------\left(4\right)$

subtracting $\left(4\right)$ from $\left(3\right)$

$$\begin{gathered} S(1-x)-xS(1-x)=1+2x+3x+4x-+........-\left(x+2x^2+3x^3+4x^4+..........\right) \\ \Rightarrow S(1-x)(1-x)=1+x+x^2+x^3+............. \end{gathered}$$

expansion of ${(1-x)}^{-1}=1+x+x^2+x^3+.....$

$S=\frac{1}{{\left(1-x\right)}^3}$

Hence, $\mathrm{Option}\left(\mathrm{D}\right)$ is correct.