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Question

$a_{n} = (- 1)^{n - 1} n^{3}; a_{9}$

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Solution

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We have, $a_{n} = (- 1)^{n - 1} n^{3}$
For, $n - 1 \sum r = 1 \frac{1}{2 - a_{r}} = \frac{2^{n - 1} (n - 1) - 1}{2^{n} - 1}$
$∴ a_{9} = {(- 1)}^{9 - 1} {(9)}^{3} = {(9)}^{3} = 729$

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