Question

Find the sum to n terms of the series:
$(x ^{2} + \frac{1}{x ^{2} } + 2) + ({x}^{4} + \frac{1}{ {x}^{4} } + 5) + ( {x}^{6} + \frac{1}{ {x}^{6} } + 8) + ...$
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Answer 1

$= > ( {x}^{2} + \frac{1}{ {x}^{2} } + 2) + ( {x}^{4} + \frac{1}{ {x}^{4} } + 5) + ( {x}^{6} + \frac{1}{ {x}^{6} } + 8) + ........... \\ = > ( {x}^{2} + {x}^{4} + {x}^{6} + ...........) + ( \frac{1}{ {x}^{2} } + \frac{1}{ {x}^{4} } + \frac{1}{ {x}^{6} } + ...........) + (2 + 5 + 8 + ...........) \\ = > \frac{ {x}^{2} ( {x}^{2n} - 1)}{ {x}^{2} - 1 } + \frac{ \frac{1}{ {x}^{2} }(1 - \frac{1}{ {x}^{2n} } ) }{1 - \frac{1}{ {x}^{2} } } + \frac{n}{2} (2 \times 2 + (n - 1)3)$

Now, simplify it and get your answer...