<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] 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
</s>