$1^2$+$2^2$-$2×3^2$+$4^2$+$5^2$-$2×6^2$+$...$+$9997^2$+$9998^2$-$2×9999^2$ ∵ For any positive integer $n$
$n^2-2(n+1)^2+(n+2)^2$
= $n^2-2n^2-4n-2+n^2+4n+4=2$
∵ $1^2$+$2^2$-$2×3^2$+$4^2$+$5^2$-$2×6^2$+$...$+$9997^2$+$9998^2$-$2×9999^2$
= $1^2+(2^2-2×3^2+4^2)+(5^2-2×6^2+7^2)+...+(9995^2-2×9996^2+9997^2)+9998^2-2×9999^2$
= $1+(2+2+...+2)+9998^2-2×9999^2$
I do not know how many lots of two will be inside the aforementioned parenthesis. 

There would be $((9995-2)/3)+1$ lots of 2. 
Following;
$1+2×(((9995-2)/3)+1)+9998^2-2×9999^2$
$2×3332+(9998^2-9999^2)+(1-9999^2)$
$6664+(9998-9999)(9998+9999)+(1-9999)(1+9999)$
$6664-19997-99980000$
= $-99993333$
