Q. The sum of $1 + n\left( 1 - \frac{1}{x} \right) + \frac{n(n + 1)}{2!}\left( 1 - \frac{1}{x} \right)^{2} + \ldots\infty$, will be

A) $x^{n}$
B) $x^{- n}$
C) $\left( 1 - \frac{1}{x} \right)^{n}$
D) None of these
Explanation

(a)
We have, ${(1 + x)}^{n} = 1 + nx + \frac{n(n - 1)}{2!}x^{2} + ...$
If $x$ is replace by $–\left( 1 - \frac{1}{x} \right)$ and $n$ is $–n$, then expression
becomes $\left\lbrack 1 - \left( 1 - \frac{1}{x} \right) \right\rbrack^{- n}$
$= 1 + ( - n)\left\lbrack - \left( 1 - \frac{1}{x} \right) \right\rbrack + \frac{( - n)( - n - 1)}{2!}\left\lbrack - \left( 1 - \frac{1}{x} \right) \right\rbrack^{2} + ...$
$\Longrightarrow x^{n} = 1 + n\left( 1 - \frac{1}{x} \right) + \frac{n(n + 1)}{2!}\left( 1 - \frac{1}{x} \right)^{2} + ...$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.