Given that x > 0, the sum ^ _ n=1 ( x x+1 )^ n-1 equals: - Vidyadip

MCQ

Given that x > 0, the sum $\sum\limits^\infty_{\text{n}=1}\Big(\frac{\text{x}}{\text{x}+1}\Big)^{\text{n}-1}$ equals:

A
$\text{x}$
✓
$\text{x}+1$
C
$\frac{\text{x}}{2\text{x}+1}$
D
$\frac{\text{x}+1}{2\text{x}+1}$

✓

Answer

Correct option: B.

$\text{x}+1$

$\sum\limits^\infty_{\text{x}+1}\Big(\frac{\text{x}}{\text{x}+1}\Big)^{(\text{n}-1)}$
$=1+\Big(\frac{\text{x}}{\text{x}+1}\Big)+\Big(\frac{\text{x}}{\text{x}+1}\Big)^2+\Big(\frac{\text{x}}{\text{x}+1}\Big)^3+\Big(\frac{\text{x}}{\text{x}+1}\Big)^4+\dots\infty$
$=\frac{1}{1-\big(\frac{\text{x}}{\text{x}+1}\big)}$ $\Big[\because$ it is a G.P. with a = 1 and $\text{r}=\Big(\frac{\text{x}}{\text{x}+1}\Big)\Big]$
$=\frac{(\text{x}+1)}{(\text{x}+1-\text{x})}$
$=\frac{(\text{x}+1)}{1}=(\text{x}+1)$

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