Find the sum of $-1^2-2^2+3^2+4^2-5^2-6^2+\cdots$ 
Find the sum of $$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2$$

By expanding the given summation,
$$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2=-1^2-2^2+3^2+4^2-5^2-6^2+\cdots+(4n-1)^2+(4n)^2$$
$$=(3^2-1^2)+(4^2-2^2)+(7^2-5^2)+(8^2-6^2)+\cdots+[(4n-1)^2-(4n-3)^2]+((4n)^2-(4n-2)^2)$$
$$=2(4)+2(6)+2(12)+2(14)+2(20)+2(22)+\cdots+2(8n-4)+2(8n-2)$$
$$=2[4+6+12+14+20+22+\cdots+(8n-4)+(8n-2)]$$
How should I proceed further?

\begin{equation*}
\begin{aligned}
\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2 &= \sum_{k=1}^{n}[((4k - 1)^2 - (4k - 3)^2) + ((4k)^2 - (4k - 2)^2)]\\
&= \sum_{k=1}^{n}[2(8k - 4) + 2(8k - 2)] \\
&= 4\sum_{k=1}^{n}[(8k - 3)] \\
&= -12n + 32\sum_{k=1}^{n}k\\
&= -12n + \frac{32n(n+1)}{2} \\
&= 16n^2 + 4n. \\
\end{aligned}
\end{equation*}
