An algebra problem by Mahtab Hossain

Rewriting the sum as $\sum _{ 1 }^{ 500 }{ 2k } -\sum _{ 1 }^{ 500 }{ 2k-1 }$ we get that

$\sum _{ 1 }^{ 500 }{ 2k } -\sum _{ 1 }^{ 500 }{ 2k-1 } =\sum _{ 1 }^{ 500 }{ 2k-(2k-1) } =\sum _{ 1 }^{ 500 }{ 1 } =500$

in every sum of two terms is 1, since there are 500 two terms in a 1000 thus 500x1 = 500

We can see that the sum of first two numbers is 1....similarly sum of next two numbers,ie.,-3 and 4 is also one...continuing to understand this algorithm we reach at a conclusion that the sum of n (if even) terms of this series is half the number of terms,ie., n/2 hence the answer is 1000/2 which gives 500. Cheers!!