# Sum of a Sequence of Odd Numbers that are Squared [duplicate]

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! :)

## marked as duplicate by Jyrki Lahtonen, Namaste discrete-mathematics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 4 '18 at 11:01

• Do you know proof by induction? – rbird Aug 4 '18 at 7:08
• I only know it very vaguely, sorry. – Cameron Choi Aug 4 '18 at 7:15
• Downvoting all "trusted" users who answer an obvious dupe. – Jyrki Lahtonen Aug 4 '18 at 8:07
• @Jyrki Lahtonen OP asked for a verification of the proof. I corrected his/her errors in bold font. – Robert Z Aug 4 '18 at 8:10
• @RobertZ You may disagree but I am not convinced by the excuse that because this user made a different error from the previous asker we should keep ten versions of an elementary calculation. – Jyrki Lahtonen Aug 4 '18 at 8:15

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 $n=(k+1)/2$.
• @CameronChoi It should be $6(2^2 + \dots + k^2) = (k+2)^3 − 2^3 − 12(k^2/4 + k/2) − 8(k/2)$ where $n=k/2$. – Robert Z Aug 4 '18 at 13:22
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 $f(m)=(2m+2)^3$
$$6\sum_{m=0}^n(2m+1)^2+2\sum_{m=0}^n1=f(n+1)-f(0)=?$$
The telescopic sum helps very well: $$\sum_{k=1}^n(2k-1)^2=\sum_{k=1}^n(4k^2-4k+1)=\sum_{k=1}^n\left(\frac{4}{3}(k^3-(k-1)^3)-\frac{1}{3}\right)=$$ $$=\frac{4}{3}(n^3-0)-\frac{n}{3}=\frac{4n^3-n}{3}.$$