Find the sum of the series: $1^{2} - 2^{2} + 3^{2} - 4^{2} + . . .$ to $n$ terms.

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Solution

$S = 1^{2} - 2^{2} + 3^{2} - 4^{2} + . . .$ to $n$ terms.
$S = 1 - 4 + 9 - 16 + 25 - 36 + . . . .$ to $n$ terms.
$S = (1 - 4) + (9 - 16) + (25 - 36) + . . . .$ to $n$ terms.
$S = - 3 - 7 - 11 - . . . .$ to $n$ terms. where $a = - 3, d = - 7 - (- 3) = - 7 + 3 = - 4$
$∴ S_{n} = \frac{n}{2} [2 \times - 3 + (n - 1) \times - 4]$
$= \frac{n}{2} [- 6 - 4 n + 4]$
$= \frac{n}{2} (- 4 n - 2)$
$= \frac{- 2 n}{2} (2 n + 1)$
$= - n (2 n + 1)$

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