Find a formula for $\sum_{i=1}^n (2i-1)^2 = 1^2+3^2+....+(2n-1)^2$ Consider the sum
$$\sum_{i=1}^n (2i-1)^2 = 1^2+3^2+...+(2n-1)^2.$$
I want to find a closed formula for this sum, however I'm not sure how to do this. I don't mind if you don't give me the answer but it would be much appreciated. I would rather have a link or anything that helps me understand to get to the answer.
EDIT: I Found this question in a calculus book so I don't really know which tag it should be.
 A: Hint: $\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}{6}$ and $\sum_{i=1}^ni=\frac{n(n+1)}{2}$. And last but not least $$(2i-1)^2=4i^2-4i+1.$$
Edit: Let's prove that $\sum_{i=1}^ni=\frac{n(n+1)}{2}$. We proceed by induction on $n$. If $n=1$ the statement is trivial. Now suppose the statement holds for $n\geq 1$. Then \begin{eqnarray}\sum_{i=1}^{n+1}i&=&\sum_{i=1}^ni+(n+1)\\
&=&\frac{n(n+1)}{2}+(n+1)\\
&=& (n+1)(\frac{n}{2}+1)\\
&=& \frac{(n+1)(n+2)}{2}.\end{eqnarray}
Here we used the induction hypothesis in the second equation. This proves the statement by induction. You can prove the other formula in a similar fashion.
A: hint
We have
$$1^2+3^2+5^2+... (2n-1)^2=$$
$=\sum $ odd$^2$=$\sum$ all$^2 $-$\sum $even$^2=$
$$\sum_{k=1}^{2n} k^2-(2^2+4^2+...4n^2)= $$
$$\sum_{k=1}^{2n}k^2-4\sum_{k=1}^n k^2=$$
$$\boxed {\color {green}{\frac {n(4n^2-1)}{3}}}$$
for $n=2$, we have $10 $ , for $n=3$, we find $35$ and for $n=4$, it is $84$.
A: Hint: Write $f(n) = \sum_{i=1}^n (2i-1)^2$. You can start by hypothesizing that the sum of squares is a cubic $f(n) = an^3 + bn^2 + cn + d$ (as a sort of discrete analogy with the integration formula $\int x^2\, dx = x^3/3 + C$). Then $(2n+1)^2 = f(n+1) - f(n) = a (3n^2 + 3n + 1) + b(2n + 1) + c$, which gets you the constants $a, b, c$, and you can find $d$ from $f(1) = 1$. Verifying this formula is an easy proof by induction.
A: $$\begin{align}
\sum_{i=1}^n (2i-1)^2
&=\sum_{i=1}^n \binom {2i-1}2+\binom {2i}2&&\text(*)\\
&=\sum_{r=0}^{2n}\binom r2\color{lightgrey}{=\sum_{r=2}^{2n}\binom r2}\\
&=\color{red}{\binom {2n+1}3}&& \text(**)\\
&\color{lightgrey}{=\frac {(2n+1)(2n)(2n-1)}6}\\
&\color{lightgrey}{=\frac 13 n(2n-1)(2n+1)}\\
&\color{lightgrey}{=\frac 13 n(4n^2-1)}
\end{align}$$

*Note that   $\;\;\;R^2=\binom R2+\binom {R+1}2$.
**Using the Hockey-stick identity
