∵ 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;
$1+2×(((9995-2)/3)+1)+9998^2-2×9999^2$
$2×3332+(9998^2-9999^2)+(1-9999^2)$
$6664+(9998-9999)(9998+9999)+(1-9999)(1+9999)$
$6664-19997-99980000$
= $-99993333$
Filling a gap for the benefit of other readers:
$$\begin{align} (n-1)^2 -2n^2 + (n+1)^2 &= (n^2-2n+1) -2n^2+(n^2+2n+1) \\ &=2n^2-2n^2+2n-2n+2 \\ &=2 \end{align}$$
or:
$$\begin{align} (n-1)^2 -2n^2 + (n+1)^2 &= ((n+1)^2-n^2) - (n^2-(n-1)^2) \\ &=(n+1+n)(n+1-n) - (n+n-1)(n-(n-1)) \\ &=(2n+1) - (2n-1) \\ &=2 \end{align}$$
or simply "the difference between squares increases by two each time"
And the accumulated number of $2$s from this will be one-third the number of terms involved in these equations, $9996/3$.
Revisiting this, we could use another grouping. The alternative pattern repeat of interest is:
$\begin{align}(n-2)^2+(n-1)^2-2n^2 &= n^2-4n+4 +n^2-2n+1 -2n^2 \\ &=-6n+5 \end{align}$
And the $n$ values here have the form $3k$, giving the full sum as
$$\begin{align} \sum_{i=1}^{3333}(5-6(3i)) &= 3333\cdot5-18 \frac{3333\cdot 3334}{2} \\ &=3333(5-30006) = 3333\times -30001 \\ &=-99993333 \end{align}$$
