<s>[INST] 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? 
 [/INST] $$\sum_{k=0}^n(2k+1)^3=\sum_{k=0}^n(2k+1+2)^3-(2n+3)^3+1$$
$$\sum_{k=0}^n(2k+1)^3=\sum_{k=0}^n[(2k+1)^3+6(2k+1)^2+12(2k+1)+8]-(2n+3)^3+1$$
$$\sum_{k=0}^n[6(2k+1)^2+12(2k+1)+8]-(2n+3)^3+1=0$$
$$\begin{align}\sum_{k=0}^n(2k+1)^2&=\frac16\left((2n+3)^3-12\sum_{k=0}^n(2k+1)-8\sum_{k=0}^n1-1\right)\\
&=\frac16\left((2n+3)^3-12\sum_{k=0}^n2k-12\sum_{k=0}^n1-8\sum_{k=0}^n1-1\right)\\
&=\frac16\left((2n+3)^3-24\sum_{k=0}^nk-20\sum_{k=0}^n1-1\right)\\
&=\frac16\left((2n+3)^3-24\sum_{k=0}^nk-20\sum_{k=0}^n1-1\right)\\
&=\frac16\left((2n+3)^3-12n(n+1)-20(n+1)-1\right)\\
&=\frac{8n^3+24n^2+22n+6}{6}\\
\end{align}$$
$$\sum_{k=0}^n(2k+1)^2=\frac{8n^3+24n^2+22n+6}{6}=\frac{(n+1)(2n+1)(2n+3)}{3}\,\,\,\,$$

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

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