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?

In the last line what you have enclosed into rectangular braces, namely $S$, can be rearranged as follows
\begin{align*}
S&=(1+3)+(2+4)+(5+7)+(6+8)+(9+11)+(10+12)+\cdots+[(4n-3)+(4n-1)]+[(4n-2)+4n]\\
&=\sum_{j=1}^{4n}j\\[3pt]
&=\frac{4n(4n+1)}{2}\\[3pt]
&=\boxed{\color{blue}{2n(4n+1)}}
\end{align*}
