# How did we derive summation term for this particular 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 $$\sumr^2$$ 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!

• $2^2 + 4^2 + \cdots + (2k)^2 = 4 \times [1^2 + 2^2 + \cdots + k^2].$ Commented Sep 18, 2021 at 14:46

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.

• Thank you very much, I appreciate this alot!
– tio
Commented Sep 18, 2021 at 15:52
• You're very welcome! :-) Commented Sep 18, 2021 at 15:53