A) \[\frac{x}{1-x}\]
B) \[\frac{1}{1-x}\]
C) \[\frac{1+x}{1-x}\]
D) 1
Correct Answer: A
Solution :
[a] The general term of the given series is \[{{t}_{n}}=\frac{{{x}^{{{2}^{n-1}}}}}{1-{{x}^{{{2}^{n}}}}}\] \[=\frac{1+{{x}^{{{2}^{n-1}}}}-1}{(1+{{x}^{{{2}^{n-1}}}})(1-{{x}^{{{2}^{n-1}}}})}\] \[=\frac{1}{1-{{x}^{{{2}^{n-1}}}}}-\frac{1}{1-{{x}^{{{2}^{n}}}}}\] Now, \[{{S}_{n}}=\sum\limits_{n=1}^{n}{{{t}_{n}}}\] \[=\left[ \begin{matrix} \left\{ \frac{1}{1-x}-\frac{1}{1-{{x}^{2}}} \right\}+\left\{ \frac{1}{1-{{x}^{2}}}-\frac{1}{1-{{x}^{4}}} \right\} \\ +...+\left\{ \frac{1}{1-{{x}^{2n-1}}}-\frac{1}{1-{{x}^{2n}}} \right\} \\ \end{matrix} \right]\] \[=\frac{1}{1-x}-\frac{1}{1-{{x}^{2n}}}\] Therefore, the sum to infinite terms is \[\underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\frac{1}{1-x}-1\] \[=\frac{x}{1-x}\]\[[\because \underset{n\to \infty }{\mathop{\lim }}\,{{x}^{2n}}=0,as\left| x \right|<1]\]
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