How do you "simplify" the sigma sign when it is raised to a power?

Question

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?

$\left(\sum\limits_{k=1}^nk\right)^2=\left(\sum\limits_{k=1}^nk\right)\cdot \left(\sum\limits_{k=1}^nk\right)$. 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? — JMoravitz, Oct 31 '16 at 05:33
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. — SirXYZ, Oct 31 '16 at 05:37
@SirJMP Thank you. So I need to show the following with induction $\sum^{n}_{k=1}k^{3} = \frac{1}{4}(n^{2} + n)^{2}$ — user376984, Oct 31 '16 at 05:43
Yes. You can also prove it by generating the required series....which Shaswata has done beautifully in his answer....and then complete using induction. — SirXYZ, Oct 31 '16 at 05:47
In general, remember that sigma notation is just shorthand for a sum $\sum_{k=1}^n a_k = (a_1 + a_2 + \dots + a_n)$, and that the usual distributive laws apply: $\left( \sum_{k=1}^n k \right)^2 = \left( \sum_{j=1}^n j \right) \cdot \left( \sum_{k=1}^n k \right) = \sum_{j=1}^n \left( j \sum_{k=1}^n k \right) = \sum_{j=1}^n \sum_{k=1}^n jk$. In this particular case, that won't help you much (it's much easier to apply the closed-form expression for $\sum_{k=1}^n k$), but it's a useful technique in general: if you can handle plain old sums, you can handle (finite) sigma expressions. — Ilmari Karonen, Oct 31 '16 at 11:12
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. — Martin Sleziak, Oct 31 '16 at 11:15
This question may be of help on the right side of the equation. And, as linked in that question, a handy table of other sums of powers. — MichaelS, Oct 31 '16 at 14:08

score 15 · Answer 1 · answered Oct 31 '16 at 05:41

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

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

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

score 6 · Answer 2 · answered Oct 31 '16 at 13:37

6

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.

answered Oct 31 '16 at 13:37

Beware of calling things coincidences. I've seen a geometric explanation of why this holds before, though I cannot recall it right now. – Paul Sinclair Oct 31 '16 at 17:59
Actually Martin Sleziak provides links to such geometric explanations in his comment on the OP. – Paul Sinclair Oct 31 '16 at 18:05
1

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. – Hypergeometricx Oct 31 '16 at 18:32

score 3 · Answer 3 · edited Oct 31 '16 at 10:36

3

$$\begin{align} \sum^{n}_{k=1}k&=\frac{n(n+1)}{2}\\ \left(\sum^{n}_{k=1}k\right)^2&=\left(\frac{n(n+1)}{2}\right)^2\\ \left(\frac{n(n+1)}{2}\right)^2&=\frac{n^2(n+1)^2}{4}\\ \sum^{n}_{k=1}k^3&=\frac{n^2(n+1)^2}{4}\\ \therefore \left(\sum^{n}_{k=1}k\right)^2&=\sum^{n}_{k=1}k^3 \end{align} $$

edited Oct 31 '16 at 10:36

Mutantoe

708

answered Oct 31 '16 at 07:05

JAP

171

How do you "simplify" the sigma sign when it is raised to a power?

3 Answers3

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