split the series into:
1^2 + 3^2 + 5^2 + 7^2 ...... 2003^2
&
-(2^2 - 4^2 + 6^2 + 8^2 ...... 2002^2)
[ ]
[ ] (2n+1) ^ 2
= summation [4n^2] + summation [1] + summation [4n]
= 4 [ (n) (n+1) (2n+1) . 1/6] + [n] + 4 [(n) (n+1) /2]
= [n^3 + 6n^2 + 10n] / 6
=
similarly for the second series
we get
- [2n^3+ 3n^2 + n]/3
adding two u should get answer
Moderation Team
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![1^2-2^2+3^2-4^2+....-2002^2+2003^2 \\ \\ \mbox{Considering in pairs}\\ \\ \ (1-2)(1+2)+(3-4)(3+4)+....(2001-2002)(2001+2002)+2003^2 \\ \\ \mbox{Taking} \ -1 \ \mbox{common in each bracket,we get} \\ \\ -1[1+2+3+....2002]+2003^2 \\ \\ = -1\left(\frac{(2002)(2003)}{2}\right)+2003^2 \\ \\ = 2003(2003-1001) = (2003)(1002) = 2007006](http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/a/b/5/ab5c8d0037fd39561cf61509c55271c6774d6cb9.gif)
