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Question: If \|x\| \< 1, then the value of 1 + n \(\left( \frac{2x}{1 + x} \right)\) + ![](https://cdn.pureess...

If |x| < 1, then the value of 1 + n (2x1+x)\left( \frac{2x}{1 + x} \right) + (2x1+x)2\left( \frac{2x}{1 + x} \right)^{2} + ........ will be –

A

(1+x1–x)n\left( \frac{1 + x}{1–x} \right)^{n}

B

(2x1+x)n\left( \frac{2x}{1 + x} \right)^{n}

C

(1+x2x)n\left( \frac{1 + x}{2x} \right)^{n}

D

(1–x1+x)n\left( \frac{1–x}{1 + x} \right)^{n}

Answer

(1+x1–x)n\left( \frac{1 + x}{1–x} \right)^{n}

Explanation

Solution

(1+x)n = nC0 + nC1x + nC2x2 ....+ nCnxn +.....)

If x is replaced by – (2x1+x)\left( \frac{2x}{1 + x} \right)and n is – n.

Hence [1–(2x1+x)]–n\left\lbrack 1–\left( \frac{2x}{1 + x} \right) \right\rbrack^{–n}

=1+(–n) [–(2x1+x)]+(–n)(–n–1)2!(–2x1+x)2+......\left\lbrack –\left( \frac{2x}{1 + x} \right) \right\rbrack + \frac{(–n)(–n–1)}{2!}\left( –\frac{2x}{1 + x} \right)^{2} + ......

= [1–x1+x]–n\left\lbrack \frac{1–x}{1 + x} \right\rbrack^{–n}= 1 +n (2x1+x)\left( \frac{2x}{1 + x} \right) + n(n+1)2\frac{n(n + 1)}{2} (2x1+x)2\left( \frac{2x}{1 + x} \right)^{2}+....

or [1+x1–x]n\left\lbrack \frac{1 + x}{1–x} \right\rbrack^{n} = 1 + n (2x1+x)\left( \frac{2x}{1 + x} \right) + n(n+1)2\frac{n(n + 1)}{2} (2x1+x)2\left( \frac{2x}{1 + x} \right)^{2}+ ..