How to prove this result related to A.P.?

Question

Questions: In an A.P. of $n$ terms ($n$ is even), the middle two terms are $p-q$ and $p+q$ respectively. Prove that the sum of the squares of all the terms of the progression is $n[p^2+\frac{n^2-1}{3}q^2]$.

My attempt: It is about a page long. It would be very tedious and difficult to post it here. I first found out the common difference and the first term of the A.P. by framing and solving two simultaneous linear equations. Then i used them to find the general term of the A.P., that is, $t_k$ and then the square of the general term, that is, $t_k^2$ . After that i used the summation operator to sum up $t_k$ from 1 to $n$. The evaluation was very cumbersome and answer also did not match.

Please help.

Answer 1

$$(p-q)^2+(p+q)^2=2(p^2+q^2)$$ $$(p-3q)^2+(p+3q)^2=2(p^2+9q^2)$$ $$\cdots$$ The sum is $$2(p^2+p^2+\cdots)+2(q^2+9q^2+25q^2+\cdots)$$ where there are $n/2$ terms in the summations.

So the final answer is $$np^2+2q^2\sum_{i=1}^{n/2}(2i-1)^2$$ $$=np^2+2q^2\sum_{i=1}^{n/2}(4i^2-4i+1)$$ $$=np^2+2q^2\left(4\frac{1}{6}\frac{n}{2}(\frac{n}{2}+1)(n+1)-4\frac{n/2(n/2+1)}{2}+\frac{n}{2}\right)$$ $$=np^2+2q^2\left(\frac{1}{6}n(n+1)(n+2)-\frac{1}{2}n(n+2)+\frac{n}{2}\right)$$ $$=np^2+2q^2\left(\frac{1}{6}n(n+2)(n+1-3)+\frac{n}{2}\right)$$ $$=np^2+2q^2\left(\frac{n}{6}(n^2-4+3)\right)$$ $$=n\left[p^2+\frac{n^2-1}{3}q^2\right]$$

Answer 2

$$\begin{align} &\sum_{r=1}^{n/2}\big(p+(2r-1)q\big)^2+\big(p-(2r-1)q\big)^2\\ &=2\sum_{r=1}^{n/2}p^2+(2r-1)^2q^2\\ &=2\cdot \frac n2 \cdot p^2+2q^2\sum_{r=1}^{n/2}(2r-1)^2\\ &=np^2+2q^2\sum_{r=1}^{n/2}\binom {2r-1}2+\binom {2r}2\\ &=np^2+2q^2\sum_{r=1}^{n}\binom r2\\ &=np^2+2q^2\binom {n+1}3\\ &=np^2+2q^2\frac{(n+1)n(n-1)}6\\ &=n\left(p^2+\frac {n^2-1}3q^2\right)\;\color{red}\blacksquare \end{align}$$