<s>[INST] Sum of $1^2+3^2+\cdots+(2n+1)^2$ Have trouble with proof. I've been working through a question on Courant's What is Mathematics?
This is the question: Prove $1^2+3^2+\cdots+(2n+1)^2=\frac{n(n+1)(2n+1)(2n+3)}{3}$. I called this $S_{(2n+1)^2}$.
What I've been given as a hint is that I can use the sum $1+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ (I called this $S_{n^2}$) in some way.
So here's my line of thought:
What do I need to do to $S_{n^2}=1^2+2^2+3^2+\cdots+n^2$ to get $S_{(2n+1)^2}=1^2+3^2+\cdots+(2n+1)^2$?
When I compared them, I thought that $S_{n^2}-(2^2+4^2+\cdots+(2n)^2) = S_{(2n+1)^2}$.
I defined $2^2+4^2+\cdots+(2n)^2$ as $S_{(2n)^2}$.
After a bit of head-scratching I managed to figure out $S_{(2n)^2}=2^2(0)+2^2(1)^2+2^2(2)^2+\cdots+2^2(n)^2=4(1^2+2^2+3^2+\cdots+n^2)=4S_{n^2} =\frac{2}{3}n(n+1)(2n+1)$
Now is where my confusion starts. So from my logic above I went forward with $S_{n^2}-S_{(2n)^2}=S_{(2n+1)^2}$ which gave me $-\frac{1}{2}(n(n+1)(2n+1))$ which isn't $S_{(2n+1)^2}$.
I don't know where I really went wrong. One of my main suspicions is that $S_{n^2}-S_{(2n)^2}=S_{n^2}-4S_{n^2}=-3S_{n^2}\neq S_{(2n+1)^2}$. But I don't see why.
 [/INST] Very easy proof
$$\sum_{k=1}^n(2k+1)^2=4\sum_{k=1}^n k^2+4\sum_{k=1}^nk+\sum_{k=1}^n1.$$
Using your hint, you can very easily conclude.
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