Question

If |x| < 1, then the sum of the 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

The correct option is D $\frac{1}{(1-x{)}^2}$

This is an A.G.P.
Let S = 1 + 2x + $3x^2+....\infty$
$\Rightarrow x.S=x+2x^2+.....\infty$
Subtracting $(1-x)S=1+x+x^2+......\infty =\frac{1}{1-x}$
$\therefore S=\frac{1}{(1-x{)}^2}.$
Aliter : Use S = $[1+\frac{r}{1-r}\times diff.ofA.P.]\frac{1}{1-r}.$