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?

Solution 1:

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 $156=12\cdot 13$.

Solution 2:

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

Solution 3:

$ =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 $

Solution 4:

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.