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This question already has an answer here:

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):

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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! :)

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marked as duplicate by Jyrki Lahtonen, Namaste discrete-mathematics Aug 4 '18 at 11:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Do you know proof by induction? $\endgroup$ – rbird Aug 4 '18 at 7:08
  • $\begingroup$ I only know it very vaguely, sorry. $\endgroup$ – Cameron Choi Aug 4 '18 at 7:15
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    $\begingroup$ Downvoting all "trusted" users who answer an obvious dupe. $\endgroup$ – Jyrki Lahtonen Aug 4 '18 at 8:07
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    $\begingroup$ @Jyrki Lahtonen OP asked for a verification of the proof. I corrected his/her errors in bold font. $\endgroup$ – Robert Z Aug 4 '18 at 8:10
  • $\begingroup$ @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. $\endgroup$ – Jyrki Lahtonen Aug 4 '18 at 8:15
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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 $n=(k+1)/2$.

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  • $\begingroup$ If I was to find the same sum, but instead with even numbers, would I write: 6(1 + 3^2 + ⋯ + k^2) = (k+2)^3 − 2 − 12((k^2)/4 + k/2) − 8((k+1)/2) $\endgroup$ – Cameron Choi Aug 4 '18 at 12:54
  • $\begingroup$ @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 $n=k/2$. $\endgroup$ – Robert Z Aug 4 '18 at 13:22
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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 $f(m)=(2m+2)^3$

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

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

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