Sequence And Series Question 104
Question: The sum of the series $ \frac{x}{1-x^{2}}+\frac{x^{2}}{1-x^{4}}+\frac{x^{4}}{1-x^{8}}+… $ to infinite terms, if $ | x |<1 $ , is
Options:
A) $ \frac{x}{1-x} $
B) $ \frac{1}{1-x} $
C) $ \frac{1+x}{1-x} $
D) 1
Show Answer
Answer:
Correct Answer: B
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}} $ $ = \begin{bmatrix} { \frac{1}{1-x}-\frac{1}{1-x^{2}} }+{ \frac{1}{1-x^{2}}-\frac{1}{1-x^{4}} } \\ +…+{ \frac{1}{1-{x^{2n-1}}}-\frac{1}{1-x^{2n}} } \\ \end{bmatrix} $ $ =\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| x |<1] $
BETA
