Find the exact closed from expression of $1^2 + 3^2 + 5^2 + · · · + (2n + 1)^ 2$ I know the above expression equals to $\frac{n(2n−1)(2n+1)}{3}$, but how exactly can i come up with something from scratch? 
 A: $$\sum_{k=0}^n(2k+1)^3=\sum_{k=0}^n(2k+1+2)^3-(2n+3)^3+1$$
$$\sum_{k=0}^n(2k+1)^3=\sum_{k=0}^n[(2k+1)^3+6(2k+1)^2+12(2k+1)+8]-(2n+3)^3+1$$
$$\sum_{k=0}^n[6(2k+1)^2+12(2k+1)+8]-(2n+3)^3+1=0$$
$$\begin{align}\sum_{k=0}^n(2k+1)^2&=\frac16\left((2n+3)^3-12\sum_{k=0}^n(2k+1)-8\sum_{k=0}^n1-1\right)\\
&=\frac16\left((2n+3)^3-12\sum_{k=0}^n2k-12\sum_{k=0}^n1-8\sum_{k=0}^n1-1\right)\\
&=\frac16\left((2n+3)^3-24\sum_{k=0}^nk-20\sum_{k=0}^n1-1\right)\\
&=\frac16\left((2n+3)^3-24\sum_{k=0}^nk-20\sum_{k=0}^n1-1\right)\\
&=\frac16\left((2n+3)^3-12n(n+1)-20(n+1)-1\right)\\
&=\frac{8n^3+24n^2+22n+6}{6}\\
\end{align}$$
$$\sum_{k=0}^n(2k+1)^2=\frac{8n^3+24n^2+22n+6}{6}=\frac{(n+1)(2n+1)(2n+3)}{3}\,\,\,\,$$

$$\sum_{k=1}^{n}(2k+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3}$$

A: $$(2k-1)^2=8\binom{k}{2}+1$$
but since
$$\sum_{k=1}^{N}\binom{k}{2}=\binom{N+1}{3}$$
it follows that:
$$\sum_{k=1}^{N}(2k-1)^2 = N+8\sum_{k=1}^{N}\binom{k}{2}=N+8\binom{N+1}{3}=\frac{1}{3}N(2N+1)(2N-1).$$
A: Most basic way in my opinion:
$$f(n)=1^2 + 3^2 + 5^2 + · · · (2n-1)^2+ (2n + 1)^ 2$$
$$f(n-1)=1^2 + 3^2 + 5^2 + · · · + (2(n -1)+ 1)^ 2=1^2 + 3^2 + 5^2 + · · · + (2n -1)^ 2$$
$$f(n)-f(n-1)=(2n + 1)^ 2$$
$$f(n)=an^3+bn^2+cn$$
$$f(n-1)=a(n-1)^3+b(n-1)^2+c(n-1)$$
$$an^3+bn^2+cn-a(n-1)^3-b(n-1)^2-c(n-1)=4n^2+4n+1$$
If you put $(n-1)^3=n^3-3n^2+3n-1$,
$(n-1)^2=n^2-2n+1$
Then you can find $a,b,c$ values easily
