Question

The sum to infinity of the series $1-3x+{5x}^2-{7x}^3+\cdots \infty$, when $|x|<1$, is

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

Answer 1

Answer: (A)

$S_{\infty }=\frac{1-x}{{(1+x)}^2}$

Let $S$ be the sum,
$S=1-3x+{5x}^2-7x^3+\dots \infty$ ...{i}
By multplyng above equation by $(-x)$
$-xS=-x+3x^2+{-5x}^3+7x^4-\dots \infty$ ...{ii}
By subtracting {ii} from {i} by shifting one place, we get,
$S(1+x)=1-2x+2x^2-{2x}^3+2x^4+\dots \infty$
By using the formula for infinite G.P.,
$a+ar+a{r}^2.....=\frac{a}{1-r}$
We get,
$S(1+x)=1-2\left(\frac{x}{1+x}\right)$
$S=\frac{1-x}{{(1+x)}^2}$