A) \[{{\left( \frac{2x}{1+x} \right)}^{n}}\]
B) \[{{\left( \frac{1+x}{2x} \right)}^{n}}\]
C) \[{{\left( \frac{1-x}{1+x} \right)}^{n}}\]
D) \[{{\left( \frac{1+x}{1-x} \right)}^{n}}\]
Correct Answer: D
Solution :
Since, \[{{(1-x)}^{-n}}=1+\frac{n}{1!}x+\frac{n(n+1)}{2!}{{x}^{2}}+....\] On putting \[x=\frac{2x}{1+x}\]both sides, we get \[{{\left( 1-\frac{2x}{1+x} \right)}^{-n}}=1+\frac{n}{1!}\left( \frac{2x}{1+x} \right)\] \[+\frac{n(n+1)}{2!}{{\left( \frac{2x}{1+x} \right)}^{2}}+.....\] \[\Rightarrow \]\[1+\frac{n}{1!}\left( \frac{2x}{1+x} \right)+\frac{n(n+1)}{2!}{{\left( \frac{2x}{1+x} \right)}^{2}}+......\] \[={{\left( \frac{1-x}{1+x} \right)}^{-n}}={{\left( \frac{1+x}{1-x} \right)}^{n}}\]
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