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?
 A: 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 $156=12\cdot 13$.
A: Hint: except $-1^2$ make pairs of other terms: $$-1^2+(-2^2+3^2)+(4^2-5^2)+(-6^2+7^2)+....$$ and then proceed further to get $-1^2+2+3+4+5+6+...$ and than solve it to get the result
A: $
=2[4+6+12+14+20+22+\cdots+(8n-4)+(8n-2)]
\\ = 4[(2+3)+(6+7)+(10+11)+\cdots+(8n-3)]
\\ = 4[5+13+21+\cdots+(8n-3)]
\\ = 4 \sum_{k=1}^n (8k-3)
\\ = 4 (4n^2+n)
\\ = 16n^2+4n
$ 
A: We know that $$\begin{align*}1+2+3+...+(4n+1)&=\frac{(4n+1) \cdot (4n+2)}{2}\\&=(2n+1)(4n+1)\end{align*}$$
So, $$\begin{align*}&2+3+(2+2)+(3+2)+6+7+...+(4n-2)+(4n-1)+(4n-2+2)+(4n-1+2)=(2n+1)(4n+1)-1 \\&\implies 2(2+3+...(4n-1))+2n \cdot 2=(2n+1)(4n+1)-1 \\&\implies 2(4+6+6+7+...+(8n-2))=2((2n+1)(4n+1)-4n-1)\\&=16n^2+4n\\&=\fbox{4n(4n+1)}\end{align*}$$
This is what you need. 
A: 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*}
A: \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*}
