Question

Evaluate $1+2x+3x^2+4x^3+.....$ upto infinity, where $\left|x\right|<1.$

Answer 1

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

Multiply both sides by $x$

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

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

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

But $(1+x+x^2+x^3+.....)$ is infinite $GP$ which is equal to $GP=\frac{a}{1-r}$

Where $a=1,r=x$

$$\begin{gathered} \Rightarrow S\left(1-x\right)=\frac{1}{1-x} \\ S=\frac{1}{{\left(1-x\right)}^2} \end{gathered}$$

Hence, $1+2x+3x^2+4x^3+.....$ upto infinity, where $\left|x\right|<1$ is $\frac{1}{{\left(1-x\right)}^2}$