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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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Given $1^2+2^2+3^2+....+2003^2$ = $(2003)(4007)(334)$ and $(1)(2003)+(2)(2002)+(3)(2001)+....+(2003)(1)$ = $(2003)(334)(x)$ then value of $x$ is


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Answer : (c) 2005
Explanation : $\;Let\;Q=1^2+2^2+....+2003^2$
answered Jan 23, 2014 by yamini.v 1 flag

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