What is the sum of the series #S=1^2-2^2+3^2-4^2+...-2002^2+2003^2#?

1 Answer
Cesareo R.
Nov 11, 2017

#2007006#

Explanation:

#sum_(k=1)^2003 (-1)^(k+1) k^2 = sum_(k=1)^1002(2k-1)^2-sum_(k=1)^1001 (2k)^2 =#

#(2*1002-1)^2+sum_(k=1)^1001((2k)^2-(2k)^2-4k+1)=#

#2003^2-sum_(k=1)^1001 (4k-1) = 2003^2-4*(1001(1001+1))/2+1001=2007006#

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