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usually the tasks look like

$$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$$

or

$$\sum_{i=0}^n i^2 = i_1^2 + i_2^2 + i_3^2+...+i_n^2$$

But for the following task I have this form:

$$\left(\sum_{k=1}^n k\right)^2 = \sum_{k=1}^nk^3 $$

First I am a little confused by how I approach this. Do I transform them into a complete term like in the first example? Or can I do it by just using the sums themselves? And how should I treat the square of the sum best?

The first step is pretty straight forward creating the base case. But as soon as I get into doing the "Induction Step" I am getting stuck and not sure how to proceed.

I am looking to know best practice for this case.

Edit: This question is a little different, since it is expected to proove this only by using complete induction using the sum notation.

Thomas Christopher Davies
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5 Answers5

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For the step you get:

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

$$[(\sum_{k=1}^n k) + (n+1)]^2 = $$

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

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

(inductive hypothesis)

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

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

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

$$\sum_{k=1}^{n+1} k^3 $$

Bram28
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  • so would you expect somebody to stay in the sumnotation doing the proof? – Thomas Christopher Davies Dec 09 '16 at 19:51
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    At first ... but then you can use the inductive hypothesis to replace the summation squared, and the single summation can be replaced by the n(n+1)/2 – Bram28 Dec 09 '16 at 19:53
  • Well I am not quite sure how our lecturer is going to evaluate this, but from seeing that highest rated answer at the top, I understand that its possible without leaving the sigma notation. If I'd go with n(n+1)/2 without having it as a theorem, I would expect it to have been proven.

    Note, we are just starting off with higher mathematics.

     – Thomas Christopher Davies Dec 09 '16 at 20:04
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    @ThomasChristopherDavies But you still need to show that last part ... for which you need the n(n+1)/2 – Bram28 Dec 09 '16 at 20:05
  • Okay, I will start working on it now. You guys just gave me the kickoff I needed to understand what to do and how to understand what is going to happen. I will close this page for now and try to do it myself. I am very thankful to this very helpful and awesome community. Thank you a lot! – Thomas Christopher Davies Dec 09 '16 at 20:08
  • Ok now I understood what you meant. Thank you a lot for the help. I worked it out :) – Thomas Christopher Davies Dec 09 '16 at 21:03
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    @ThomasChristopherDavies Cool! – Bram28 Dec 09 '16 at 21:32
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Assume that $\displaystyle\left(\sum_{k=1}^n k\right)^2 = \sum_{k=1}^nk^3$ holds for $n.$ We want to show that $\displaystyle\left(\sum_{k=1}^{n+1} k\right)^2 = \sum_{k=1}^{n+1}k^3.$ How to do it? Note that

$$\begin{align}\left(\sum_{k=1}^{n+1} k\right)^2&=\left(\sum_{k=1}^{n} k+n+1\right)^2\\&= \color{blue}{\left(\sum_{k=1}^{n} k\right)^2}+2(n+1)\sum_{k=1}^nk+(n+1)^2\\&\underbrace{=}_{\rm{induction}\:\rm{hypothesis}}\color{blue}{\sum_{k=1}^nk^3}+\color{red}{2(n+1)\sum_{k=1}^nk+(n+1)^2}\\&=\sum_{k=1}^{n+1}k^3\end{align}$$ if and only if $\displaystyle(n+1)^3=2(n+1)\sum_{k=1}^nk+(n+1)^2.$ Show this equality and you are done.

mfl
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You could write $$\left( \sum_{k=1}^{n+1} k \right)^2 = \left[ \left(\sum_{k=1}^n k \right)+ (n+1) \right]^2 = \left(\sum_{k=1}^n k \right)^2 + 2(n+1) \left( \sum_{k=1}^n k\right) + (n+1)^2$$ The induction step assumes that the first term on the right is just $\displaystyle \sum_{k=1}^n k^3$. You need to show that the rest of the terms add up to $(n+1)^3$.

Umberto P.
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I'd do this is using induction to prove that $\sum_{k=1}^n k = \frac{n}{2}(n+1)$ and $\sum_{k=1}^n k^3 = \frac{n^2}{4}(n+1)^2$; these are straightforward induction proofs.

Then $$\left[\frac{n}{2}(n+1)\right]^2 = \frac{n^2}{4}(n+1)^2 \iff \left(\sum_{k=1}^n k\right)^2 = \sum_{k=1}^n k^3$$

Zain Patel
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  • Well, can i just use the n/2(n+1) as axiom though they are not given? – Thomas Christopher Davies Dec 09 '16 at 20:10
  • Perhaps, it's not too hard to prove without induction (just add the first n natural numbers to themselves in reverse and divide by 2) but I'd include it for the sake of completeness. (and it wouldn't be classified as an axiom, just a standard result) – Zain Patel Dec 09 '16 at 20:12
  • well the task expects me to prove it my complete induction, so I guess I got no choice here, but its a good mindbump to think about it :) – Thomas Christopher Davies Dec 09 '16 at 20:18
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I know you have to use induction, so this doesn't help you, but below is a cool proof by picture:

enter image description here

Bram28
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