Find the exact closed from expression of $1^2 + 3^2 + 5^2 + · · · + (2n + 1)^ 2$

Question

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?

Answer 1

$$\sum_{k=0}^n(2k+1)^3=\sum_{k=0}^n(2k+1+2)^3-(2n+3)^3+1$$ $$\sum_{k=0}^n(2k+1)^3=\sum_{k=0}^n[(2k+1)^3+6(2k+1)^2+12(2k+1)+8]-(2n+3)^3+1$$

$$\sum_{k=0}^n[6(2k+1)^2+12(2k+1)+8]-(2n+3)^3+1=0$$

$$\begin{align}\sum_{k=0}^n(2k+1)^2&=\frac16\left((2n+3)^3-12\sum_{k=0}^n(2k+1)-8\sum_{k=0}^n1-1\right)\\ &=\frac16\left((2n+3)^3-12\sum_{k=0}^n2k-12\sum_{k=0}^n1-8\sum_{k=0}^n1-1\right)\\ &=\frac16\left((2n+3)^3-24\sum_{k=0}^nk-20\sum_{k=0}^n1-1\right)\\ &=\frac16\left((2n+3)^3-24\sum_{k=0}^nk-20\sum_{k=0}^n1-1\right)\\ &=\frac16\left((2n+3)^3-12n(n+1)-20(n+1)-1\right)\\ &=\frac{8n^3+24n^2+22n+6}{6}\\ \end{align}$$

$$\sum_{k=0}^n(2k+1)^2=\frac{8n^3+24n^2+22n+6}{6}=\frac{(n+1)(2n+1)(2n+3)}{3}\,\,\,\,$$

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

Answer 2

$$(2k-1)^2=8\binom{k}{2}+1$$ but since $$\sum_{k=1}^{N}\binom{k}{2}=\binom{N+1}{3}$$ it follows that: $$\sum_{k=1}^{N}(2k-1)^2 = N+8\sum_{k=1}^{N}\binom{k}{2}=N+8\binom{N+1}{3}=\frac{1}{3}N(2N+1)(2N-1).$$

Answer 3

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