Sum of a Sequence of Odd Numbers that are Squared

Question

What is the sum of all the numbers in the sequence $1^2 + 3^2 + 5^2 + 7^2 + 9^2 + \ldots + k^2$. Note that all the numbers being squared in the sequence are all odd numbers.

This is what I have done so far (sorry if the images are an inconvenience, but this was the clearest way to display my working out):

I am a little stuck on what to do next and how to obtain $\frac{n (4n^2 - 1)}{3}$ as the final result as this is what I am meant to end up with. It would be really appreciated if anyone could make suggestions towards completing and improving my method. Thanks! :)

Answer 1

Your approach is almost correct. Check again your steps. At the end you should have $$\begin{align}(k+\mathbf{2})^3-1 &=6(1+3^2+\dots+k^2)+12(1+3+\dots+k)+\underbrace{(8+8+\dots+8)}_{\text{$(k+1)/2$ times}}\\ &=6(1+3^2+\dots+k^2)+12\left(\frac{k+1}{2}\right)^2+8\mathbf{\left(\frac{k+1}{2}\right)}.\end{align}$$ Hence $$6(1+3^2+\dots+k^2)=(k+\mathbf{2})^3-1-12\left(\frac{k+1}{2}\right)^2-8\mathbf{\left(\frac{k+1}{2}\right)}$$ and it follows that $$\sum_{j=1}^n(2j-1)^2=1+3^2+\dots+k^2=\frac{k(k+2)(k+1)}{6}=\frac{n (4n^2 - 1)}{3}$$ where $n=(k+1)/2$.

Answer 2

Set $n=2m$ $$(2m+2)^3-(2m)^3=24m^2+24m+8=6(2m+1)^2+2$$

$$\implies\sum_{m=0}^n(6(2m+1)^2+2)=\sum_{m=0}^n((2m+2)^3-(2m)^3)=\sum_{m=0}^n(f(m+1)-f(m))$$ where $f(m)=(2m+2)^3$

$$6\sum_{m=0}^n(2m+1)^2+2\sum_{m=0}^n1=f(n+1)-f(0)=?$$

Answer 3

The telescopic sum helps very well: $$\sum_{k=1}^n(2k-1)^2=\sum_{k=1}^n(4k^2-4k+1)=\sum_{k=1}^n\left(\frac{4}{3}(k^3-(k-1)^3)-\frac{1}{3}\right)=$$ $$=\frac{4}{3}(n^3-0)-\frac{n}{3}=\frac{4n^3-n}{3}.$$