I am currently learning about arithmetic series and sequences, and I am stuck on this question. I have no idea how to proceed, and I cannot find any way to simplify the expression given.
Here is the question:
Compute the sum$ (a +(2n+1)d)^2- (a + (2n)d)^2 +(a + (2n-1)d)^2 - (a+(2n-2)d)^2 + \cdots + (a+d)^2 - a^2.$
We need to evaluate
$$\sum_{k=0}^{2n+1} (-1)^{k}(a+(2n+1-k)d)^2=\sum_{k=0}^{N} (-1)^{k}(a+(N-k)d)^2$$
with
$$(a+(N-k)d)^2=(a^2+2adN+d^2N^2)-(2ad+2d^2N)\,k+d^2\,k^2$$
therefore we can evaluate the original sum from the three sums
$$\sum_0^N (-1)^k \quad \sum_0^N (-1)^k k\quad \sum_0^N (-1)^kk^2$$
To compute this sums, starting for the ordinary not alternating sums, we can use the following
$$\sum_{k=0}^{2n+1}(-1)^k a_k =\sum_{k=0}^{2n+1} a_k-2 \sum_{k=0}^{n} a_{2k}=\sum_{k=0}^{n} a_{2k+1}- \sum_{k=0}^{n} a_{2k}$$
Refer also to the related
$\sum_{k=0}^{n} -(a+2kd)^2 + \sum_{k=0}^{n} (a+(2k+1)d)^2= \sum_{k=0}^{n} -a^2 -4kad -(2kd)^2 +a^2 +4kad+2ad+((2k)^2+4k+1)d^2= \sum_{k=0}^{n} 2ad+ 4kd^2+d^2=(2ad+d^2)\sum_{k=0}^{n}+4d^2\sum_{k=0}^{n} k$
Can you take it from here?
You can write your expression as
$ = \sum \limits_{i = 0}^{n} (a +(2n+1-2i)d)^2- (a + (2n-2i)d)^2 $
$= \sum \limits_{i = 0}^{n}2ad + [4(n-i)+1]d^2$
$= 2(n+1)ad + (4n+1)(n+1)d^2 - 2 n(n+1) d^2$
$= 2(n+1)ad + (2n+1)(n+1)d^2$
