Question

If x is positive, the sum of infinity of the series $\frac{1}{1+x}-\frac{1-x}{(1+x{)}^2}+\frac{(1-x{)}^2}{(1+x{)}^2}-\frac{(1-x{)}^3}{(1+x{)}^4}+....$ is

• A. $\frac{1}{2}$
• B. $\frac{3}{4}$
• C. 1
• D. none of these

Answer 1

The correct option is A

$\frac{1}{2}$

(a) $\frac{1}{2}$

Let $S=\frac{1}{(1+x)}-\frac{(1-x)}{(1+x{)}^2}+\frac{(1-x{)}^2}{(1+x{)}^3}-\frac{(1-x{)}^3}{(1+x{)}^4}+....\infty$

It is clear that it is a G.P. with $a=\frac{1}{(1+x)}$ and

$r=-\frac{(1-x)}{(1+x)}$

$\therefore S=\frac{a}{(1-r)}$

$\Rightarrow S=\frac{\frac{1}{(1+x)}}{[1-(-\frac{(1-x)}{(1+x)})]}$

$\Rightarrow S=\frac{\frac{1}{(1+x)}}{[1+(-\frac{1-x}{1+x})]}$

$\Rightarrow S=\frac{\frac{1}{(1+x)}}{\left[\frac{(1+x)(1-x)}{(1+x)}\right]}$

$\Rightarrow S=\frac{1}{2}$