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

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Solution

We want to find $1^{2} - 2^{2} + 3^{2} - 4^{2} + . . . .$ to $2 n$ terms
$= 1 - 4 + 9 - 16 + 25 - . . . .$ to $2 n$ terms
$= (1 - 4) + (9 - 16) + (25 - 36) + . . . .$ to $n$ terms. (after grouping)
$= - 3 + (- 7) + (- 11) + . . . . n$ terms
Now, the above series is in an A.P. with first term $a = - 3$ and common difference $d = - 4$
Therefore, the required sum $= \frac{n}{2} [2 a + (n - 1) d]$
$= \frac{n}{2} [2 (- 3) + (n - 1) (- 4)]$
$= \frac{n}{2} [- 6 - 4 n + 4] = \frac{n}{2} [- 4 n - 2]$
$= \frac{- 2 n}{2} (2 n + 1) = - n (2 n + 1)$ .

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