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Question
Given that x > 0, the sum \[\sum^\infty_{n = 1} \left( \frac{x}{x + 1} \right)^{n - 1}\] equals
Options
(a) x
(b) x + 1
(c) \[\frac{x}{2x + 1}\]
(d) \[\frac{x + 1}{2x + 1}\]
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Solution
(b) x + 1
\[\sum^\infty_{n = 1} \left( \frac{x}{x + 1} \right)^\left( n - 1 \right) = 1 + \left( \frac{x}{x + 1} \right) + \left( \frac{x}{x + 1} \right)^2 + \left( \frac{x}{x + 1} \right)^3 + \left( \frac{x}{x + 1} \right)^4 + . . . \infty \]
\[ = \frac{1}{1 - \left( \frac{x}{x + 1} \right)} \left[ \because \text{ it is a G . P } . \text{ with a = 1 and } r = \left( \frac{x}{x + 1} \right) \right]\]
\[ = \frac{\left( x + 1 \right)}{\left( x + 1 - x \right)}\]
\[ = \frac{\left( x + 1 \right)}{1} = \left( x + 1 \right)\]
\[\]
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