If $S = 1^{2} - 2^{2} + 3^{2} - 4^{2}$ up to $n$ terms if $n$ is even, then S equals

\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 C $\frac{- n (n + 1)}{2}$
As $n$ is even, so the given series can be written as
$S = (1^{2} - 2^{2}) + (3^{2} - 4^{2}) + (5^{2} - 6^{2}) + . . . . + ((n - 1)^{2} - n^{2})$
$S = (1 - 2) (1 + 2) + (3 - 4) (3 + 4) + (5 - 6) (5 + 6) + . . . + ((n - 1) - n) ((n - 1) + n)$

$S = - [1 + 2 + 3 + 4 + . . . . + (n - 1) + n]$
$= - \frac{n (n + 1)}{2}$

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Similar questions

1 + \frac{1 + 2}{2} + \frac{1 + 2 + 3}{3} + . . . .

\frac{n (n + 3)}{4}

\frac{n (n + 2)}{4}

\frac{n (n + 1) (n + 2)}{6}

1^{2} - 2^{2} + 3^{2} - 4^{2} + 5^{2} - 6^{2} + \dots

n

\frac{1^{2}}{1} + \frac{1^{2} + 2^{2}}{1 + 2} + \frac{1^{2} + 2^{2} + 3^{2}}{1 + 2 + 3} + \dots

If 1+ $\frac{1 + 2}{2} + \frac{1 + 2 + 3}{3} + . . . .$ to n terms is S. Then, S is equal to

\frac{1 \times 2^{2} + 2 \times 3^{2} + 3 \times 4^{2} + \dots + n (n + 1)^{2}}{1^{2} \times 2 + 2^{2} \times 3 + 3^{2} \times 4 + \dots + n^{2} (n + 1)} = \frac{8}{7}

n =

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