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Question
If x is positive, the sum to infinity of the series \[\frac{1}{1 + x} - \frac{1 - x}{(1 + x )^2} + \frac{(1 - x )^2}{(1 + x )^3} - \frac{(1 - x )^3}{(1 + x )^4} + . . . . . . is\]
Options
(a) 1/2
(b) 3/4
(c) 1
(d) none of these
Solution
(a) \[\frac{1}{2}\]
\[\text{ Let } S = \frac{1}{\left( 1 + x \right)} - \frac{\left( 1 - x \right)}{\left( 1 + x \right)^2} + \frac{\left( 1 - x \right)^2}{\left( 1 + x \right)^3} - \frac{\left( 1 - x \right)^3}{\left( 1 + x \right)^4} + . . . \infty \]
\[\text{ It is clear that it is a G . P . with a } = \frac{1}{\left( 1 + x \right)} \text{ and }r = - \frac{\left( 1 - x \right)}{\left( 1 + x \right)} . \]
\[ \therefore S = \frac{a}{\left( 1 - r \right)}\]
\[ \Rightarrow S = \frac{\frac{1}{\left( 1 + x \right)}}{\left[ 1 - \left( - \frac{\left( 1 - x \right)}{\left( 1 + x \right)} \right) \right]}\]
\[ \Rightarrow S = \frac{\frac{1}{\left( 1 + x \right)}}{\left[ 1 + \frac{\left( 1 - x \right)}{\left( 1 + x \right)} \right]}\]
\[ \Rightarrow S = \frac{\frac{1}{\left( 1 + x \right)}}{\left[ \frac{\left( 1 + x \right) + \left( 1 - x \right)}{\left( 1 + x \right)} \right]}\]
\[ \Rightarrow S = \frac{1}{2}\]
\[\]
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