The value of
$1^{2} - 2^{2} + 3^{2} - 4^{2} + 5^{2} - 6^{2} + . . . . + 99^{2} - 100^{2}$ is

- 100

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- 5050

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- 2500

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- 2520

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Solution

The correct option is D $- 5050$
Here, using the formula
$a^{2} - b^{2} = (a + b) * (a - b)$
Since, $a$ and $b$ will be consecutive numbers and $a < b ⟹$ difference will be $- 1$
$∴ (a^{2} - b^{2}) = - (a + b)$
$∴$ Value of the expression
$1^{2} - 2^{2} + 3^{2} - 4^{2} + . . . . . + 99^{2} - 100^{2}$
$= - (1 + 2 + 3 + 4 + 5 - - - - - - - - - - - - - 100)$
Sum of $n$ consecutive numbers $= \frac{n (n + 1)}{2}$
Then, we have
$1^{2} - 2^{2} + 3^{2} - 4^{2} + . . . . . + 99^{2} - 100^{2}$
$= - (\frac{100 (101)}{2}) = - 5050$

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