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How do you simplify the following expression:

$$\left(\sum^{n}_{k=1}k \right)^2$$

I am supposed to show that

$$\left(\sum^{n}_{k=1}k \right)^2 = \sum^{n}_{k=1}k^{3} $$

The problem is I do not really know how to manipulate the sigma sign. I know that I (probably) need to use induction somehow, but the main question is how do you "simplify" the sigma sign when it is raised to a power. Due to the problem itself I know that (most likely); $$\left(\sum^{n}_{k=1}k \right)^2 = \sum^{n}_{k=1}k^{3} $$ so is it possible to simply manipulate the LHS so that it looks like the RHS?

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    $\begingroup$ $\left(\sum\limits_{k=1}^nk\right)^2=\left(\sum\limits_{k=1}^nk\right)\cdot \left(\sum\limits_{k=1}^nk\right)$. Now, what do you know about $\left(\sum\limits_{k=1}^nk\right)$? Do you know a nice convenient expression for it? If you don't, can you come up with a conjecture after looking at the pattern and prove it via induction? Pictures might help for a geometric interpretation for this step. Now, can you continue via induction for the overall claim? $\endgroup$
    – JMoravitz
    Oct 31, 2016 at 5:33
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    $\begingroup$ You need to get a single expression for the sum represented by the sigma sign.....which in this case is equal to $n(n+1)/2$. Then you can square this result to obtain the final answer. $\endgroup$
    – SirXYZ
    Oct 31, 2016 at 5:37
  • $\begingroup$ @SirJMP Thank you. So I need to show the following with induction $\sum^{n}_{k=1}k^{3} = \frac{1}{4}(n^{2} + n)^{2}$ $\endgroup$
    – user376984
    Oct 31, 2016 at 5:43
  • $\begingroup$ Yes. You can also prove it by generating the required series....which Shaswata has done beautifully in his answer....and then complete using induction. $\endgroup$
    – SirXYZ
    Oct 31, 2016 at 5:47
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    $\begingroup$ See also the following posts: Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction and other posts linked there. Proving the identity $\sum_{k=1}^n {k^3} = \big(\sum_{k=1}^n k\big)^2$ without induction and other posts linked there. $\endgroup$
    – Martin Sleziak
    Oct 31, 2016 at 11:15

3 Answers 3

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

$$=(n+1)\left(\frac{(n+1)(n+2)}{2}+\frac{(n+1)(n)}{2}\right)=(n+1)^3$$

I guess you can do the rest now since you have already figured out that you need to use induction.

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This identity is a coincidence — it is not proven by doing general series manipulations, but instead by simply computing the left and right hand sides and confirming they're equal.

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  • $\begingroup$ Beware of calling things coincidences. I've seen a geometric explanation of why this holds before, though I cannot recall it right now. $\endgroup$
    – Paul Sinclair
    Oct 31, 2016 at 17:59
  • $\begingroup$ Actually Martin Sleziak provides links to such geometric explanations in his comment on the OP. $\endgroup$
    – Paul Sinclair
    Oct 31, 2016 at 18:05
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    $\begingroup$ It appears that the point (a very pertinent one (+1)) by made in this solution is that, in general, the square of a summation cannot be manipulated into a single neat summation, and that the identity in the question is probably a special case. $\endgroup$
    – Hypergeometricx
    Oct 31, 2016 at 18:32
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$$\begin{align} \sum^{n}_{k=1}k&=\frac{n(n+1)}{2}\\ \left(\sum^{n}_{k=1}k\right)^2&=\left(\frac{n(n+1)}{2}\right)^2\\ \left(\frac{n(n+1)}{2}\right)^2&=\frac{n^2(n+1)^2}{4}\\ \sum^{n}_{k=1}k^3&=\frac{n^2(n+1)^2}{4}\\ \therefore \left(\sum^{n}_{k=1}k\right)^2&=\sum^{n}_{k=1}k^3 \end{align} $$

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