Question

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

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

Answer 1

Answer: (D)

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

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

multiply by $x$ on both sides

$xS=x+2x^2+3x^3+.....\infty$ $\dots \left(2\right)$

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

$S=1+2x+3x^2+......\infty$

$xS=x+2x^2+3x^3+.....\infty$

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$S(1-x)=1+x+x^2+x^3+.....\infty$

$S(1-x)=\frac{1}{1-x}$ [sum of infinte G.P $S_{\infty }=\frac{1}{1-r}]$

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