Sum of the series
S = $1^{2} - 2^{2} + 3^{3} - 4^{2} + . . . . - 2002^{2} + 2003^{2}$ is

2007006

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1005004

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2000506

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none of these

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Solution

The correct option is A 2007006
$S = 1^{2} - 2^{2} + 3^{2} - 4^{2} + . . . - 2002^{2} + 2003^{2}$

$= (1^{2} - 2^{2}) + (3^{2} - 4^{2}) + (5^{2} - 6^{2}) + (7^{2} - 8^{2}) + . . . + (2001^{2} - 2002^{2}) + 2003^{2}$

$= (1 - 2) (1 + 2) + (3 - 4) (3 + 4) + (5 - 6) (5 + 6) + . . + (2001 - 2002) (2001 + 2002) + 2003^{2}$

$= (- 1) (1 + 2) + (- 1) (3 + 4) + (- 1) (5 + 6) + . . + (- 1) (2001 + 2002) + 2003^{2}$

$= - 1 - 2 - 3 - 4 - 5 - 6 - . . - 2001 - 2002 + 2003^{2}$

$= - (1 + 2 + 3 + 4 + 5 + 6 + . . . + 2001 + 2002) + 2003^{2}$ where sum of natural numbers $= \frac{n (n + 1)}{2}$

$= - \frac{2002 (2002 + 1)}{2} + 2003^{2}$

$= - 1001 (2003) + 2003^{2}$

$= 2003 (- 1001 + 2003)$

$= 2003 (1002)$

$= 2003 (1000 + 2)$

$= 2003000 + 4006 = 2007006$

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