A sum of series problem with alternating sign of terms

Question

I came across a problem that requires me to find the sum of a series. The term of the series $T_n$ is given by $$T_n = (-1)^{\frac{n(n+1)}2}n^2$$

Sum till $4n$ terms is to be found.

Writing down the first few terms:

$$-(1)^2-(2)^2+3^2+4^2-(5)^2-(6)^2+7^2+8^2+ \ldots$$

This can be arranged into a sum of difference between squares:

$$(3^2-1^2)+(4^2-2^2)+(7^2-5^2)+(8^2-6^2)+\ldots$$ $$=2\times(4+6+12+14+20+\ldots)$$ $$=4\times(2+3+6+7+10\ldots)$$

In the last step, the expression inside the parentheses contains $n$ terms.

I am unable to proceed from here. How do I find the sum of the series inside the parentheses? A hint would be much appreciated.

This is a question from Resonance DLPD Algebra, Page #30, Question #2.

Answer 1

If the number of terms is multiple of$4,$ $$\sum_{r=0}^n(-(4r+1)^2-(4r+2)^2+(4r+3)^2+(4r+4)^2)=2\sum(8r+4+8r+6)=?$$

Answer 2

$$\begin{align} &\;\;\;\ \sum_{r=1}^{4n}(-1)^{\frac {r(r+1)}2}r^2\\ &=\sum_{r=1}^n (4r)^2+(4r-1)^2-(4r-2)^2-(4r-3)^2\\ &=\sum_{r=1}^n \underbrace{(4r)^2-(4r-2)^2}+\underbrace{(4r-1)^2-(4r-3)^2}\\ &=\sum_{r=1}^n \qquad 2(8r-2)\;\;\;+\qquad 2(8r-4)\\ &=\sum_{r=1}^n4(8r-3)\\ &=4\sum_{r=1}^n (8r-4+1)\\ &=4\sum_{r=1}^n 4(2r-1)+1\\\\ &=4(4n^2+n)\\\\ &=4n(4n+1)\end{align}$$