If the general pair term is n^2-(n+1)^2 we can factorise the pair: (n-n-1)(n+n+1)=-(2n+1), where n increases in steps of 2 (1, 3, 5, etc.). We can write n=2p-1 where p≥1, so -(2n+1)=-(4p-1).
This gives us the series: -3-7-11-15-...-4019-4023+2013^2. -4023=-4p+1, so p=4024/4=1006 at the limit.
We can take pairs of terms: -(3+4023)-(7+4019)-(11+4015)-...+2013^2.
-4026m+2013^2 where m=max p/2=1006/2=503.
The sum of the series is 2013^2-4026*503=2027091.