If $n$ is odd, then the sum of $n$ terms of the series $1^{2} - 2^{2} + 3^{2} - 4^{2} + 5^{2} - 6^{2} + . . .$ is

\frac{n (n - 1)}{2}

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\frac{n (n + 1)}{2}

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\frac{- n (n - 1)}{2}

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\frac{- n (n + 1)}{2}

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Solution

The correct option is B $\frac{n (n + 1)}{2}$
If $n$ is odd, then required sum is given by
$1^{2} - 2^{2} + 3^{2} - 4^{2} + 5^{2} - 6^{2} + . . . . . + {(n - 2)}^{2} - {(n - 1)}^{2} + n^{2}$
$= - (1 + 2 + 3 + . . . . + (n - 2) + (n - 1) + n^{2})$
$= - \frac{n (n - 1)}{2} + n^{2} = \frac{n (n + 1)}{2}$

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