IIT JEE , AIEEE Forum - Mathematics , Algebra

jic

Find the sum of the series

1^2 – 2^2 + 3^2 - …………….- 2002^2 + 2003^2

anchitsaini

Expand it as

1^2 + 2^2 + 3^2 - ?????.+ 2002^2 + 2003^2
-2[2^2 + 4^2 + 6^2 - ?????.+ 2002^2 ]

=1^2 + 2^2 + 3^2 - ?????.+ 2002^2 + 2003^2
-2 * 2^2[1^2 + 2^2 + 3^2 + .... 1001^2]

oneyeartogo

I think this has been asked before:

${1}^{2}-{2}^{2}+{3}^{2}..............-{2002}^{2} + {2003}^{2}\\=> (1-2) (1+2) + (3-4)(3+4)................(2001-2002)(2001+2002) +{2003}^{2}\\=> (-1) (1+2+3+4..........+2002) + {2003}^{2}\\=> (2003)(1002)\\=>2007006$

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