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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! :)

Michael Rozenberg
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Cameron Choi
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  • Do you know proof by induction? – A. Goodier Aug 04 '18 at 07:08
  • I only know it very vaguely, sorry. – Cameron Choi Aug 04 '18 at 07:15
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    Downvoting all "trusted" users who answer an obvious dupe. – Jyrki Lahtonen Aug 04 '18 at 08:07
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    @Jyrki Lahtonen OP asked for a verification of the proof. I corrected his/her errors in bold font. – Robert Z Aug 04 '18 at 08:10
  • @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 04 '18 at 08:15
  • @Jyrki Lahtonen So, in your opinion, I answered just to gain some points. – Robert Z Aug 04 '18 at 08:22
  • @RobertZ The real problem here is that having several variants of the same question makes it more difficult to use the site. Answering these questions A) increases the entropy, B) creates a culture that the askers need not search, C) invites more minor variants of the same question. I want to disincentivize that kind of behavior. Downvoting is the only tool I have. Commenting also to make the reason of the downvote clear. – Jyrki Lahtonen Aug 04 '18 at 08:26
  • But, an extended discussion of how the site really should be used should take place in meta. We do have a faction of users operating on the principle will do your homework for 10 ego points. I don't know whether you personally belong to that faction. You see, the prolific answerers rarely come to meta to justify their way of using the site even though many find it actuvely harmful. – Jyrki Lahtonen Aug 04 '18 at 08:29
  • Mind you, I upvoted the quesion as A) the asker showed their effort, B) as a new user should be given some slack about not knowing the site norms. I don't think it is asking too much that any experienced user should know that "there's snowball's chance in hell that this sum has not been handled before". I think the proper way would be to 1) point out the mistake in a comment, 2) locate a copy, and vote to close this as a dupe, 3) if you have a novel way of arriving at this formula post it as an answer to the original. – Jyrki Lahtonen Aug 04 '18 at 08:33
  • @RobertZ I do see the point of you addressing the mistake rather than just posting your own solution. Will retract my downvote to you. I need to do a fake edit. Sorry about that. And sorry about missing this subtlety. Thanks for bringing it to my notice. I was seeing red :-( – Jyrki Lahtonen Aug 04 '18 at 08:36
  • 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) – Cameron Choi Aug 04 '18 at 12:56

3 Answers3

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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 $n=(k+1)/2$.

Jyrki Lahtonen
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Robert Z
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  • 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) – Cameron Choi Aug 04 '18 at 12:54
  • @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 04 '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 $f(m)=(2m+2)^3$

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

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

Michael Rozenberg
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