The sum of the series $1^{2} - 2^{2} + 3^{2} - 4^{2} + . . . . . . . . . . - 1000^{2} + 1001^{2}$ is

- 501 \times 1001

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501 \times 1001

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499 \times 1001

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- 499 \times 1001

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Solution

The correct option is A $501 \times 1001$
We would like to regroup the series as follows:
$1^{2} + (3^{2} - 2^{2}) + (5^{2} - 4^{2}) + . . . . (1001^{2} - 1000^{2})$
$a^{2} - b^{2} = (a + b) (a - b)$ , in our case $a - b = 1$
Hence,
$1 + 5 + 9 + 13 + . . . + 2001$
There are $501$ terms, then
Sum $= 501 \times 1001$

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