Find the sum of $n$ terms of the series whose $n t h$ term is :
$(2 n - 1)^{2}$

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Solution

$\sum (2 n - 1)^{2}$
$\sum 4 n^{2} - 4 n + 1$
$4 \sum n^{2} - 4 \sum n + \sum 1$
$\frac{4 n (n + 1) (2 n + 1)}{6} + \frac{4 n (n + 1)}{2} + n$

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