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?

HINT: Split your last sum into the sum of two arithmetic progressions, each of length $n$.
An alternative is to calculate a few values of the sum, guess a closed form, and then prove the closed form. For $n=1,2,3$ the sum in question is $20,72,156$, respectively. Note that $20=4\cdot5$, $72=8\cdot9$, and $156=12\cdot 13$.
