How did we derive summation term for this particular question?

Question

I've been self-studying summation of series and there's this one question that I can't seem to completely understand.

The question is:

Find the sum of the series

$1^2$ + $2$ x $2^2$ + $3^2$ + $2$ x $4^2$ +..... $2(n-1)^2$ + $n^2$

where n is odd

Here's what I did:

I broke the series into two parts so that I had $n^2$ and then $2(n-1)^2$ and applied the summation formula

For $n^2$, I got $\sum$$r^2$ and for $2(n-1)^2$, I expanded it first and got my final summation as $2$$\sum$ $r^2$ - $2r$ + $1$

In the working done in my book, however, they do break the series into two parts but for $2(n-1)^2$ they use the sum of square formulas and come up with the term $\frac{4\left(\frac{n-1}{2}\right)\left(\frac{n+1}{2}\right)n}{6}$

Can anyone please explain or give me a hint as to how they came up with that formula?

I've been stuck on this for quite some time and I would greatly appreciate any help!

Answer 1

We know that $n$ is odd, so let $n=2k+1$ for some integer $k$.

$$\sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}{6}$$

and

$$2^{2}+4^{2}+6^{2}+...+(n-1)^2=2^{2}+4^{2}+6^{2}+...+(2k)^{2}$$ $$=2^{2}+2^{2}2^{2}+2^{2}3^{2}+...+2^{2}k^2$$ $$=2^{2}\left(1^{2}+2^{2}+3^{2}+...+k^{2}\right)=2^{2}\frac{k(k+1)(2k+1)}{6}=\frac{2k(k+1)(2k+1)}{3}=\frac{2\left(\frac{n-1}{2}\right)\left(\frac{n+1}{2}\right)n}{3}$$

since $k=\frac{n-1}{2}$. Then we have

$$1^{2}+3^{2}+5^{2}+...+n^2=1^2+3^2+5^2+...+(2k+1)^2$$ $$=1^{2}+2^{2}+3^{2}+....+(2k+1)^2-\left[2^{2}+4^{2}+...+(2k)^{2}\right]$$ $$=\frac{(2k+1)(k+1)(4k+3)}{3}-\frac{2k(k+1)(2k+1)}{3}$$ $$=\frac{(k+1)(2k+1)(2k+3)}{3}$$

Now let express this in terms of $n$ and combine the two expression to obtain your final result.