# Find the closed form of a summation of sums

So I'm trying to find a closed form for a series (going from 1 to n, for any n > 0). I'm a computer science student only, so my mathematical background is limited in this case.

Anywho, here is how far I've managed to get:

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

Getting to this point took me a little over two hours, but now I've been staring at this and really don't know where to go from here. Any help is appreciated. Thanks! :)

PS: I notice the pattern that appears => 1(1) + 2(1+2) + 3(1+2+3) + ... + n(1+2+3+...+n), but I'm not sure what to extrapolate from there on.

Edit: removed the equals sign. It probably shouldn't have been there in the beginning.

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It might interest you that $$\sum_{i=1}^n i^3 = \left ( \sum_{i=1}^n i\right )^2$$ – Ron Gordon Jan 22 '14 at 20:49
What's the question here, anyway?? – DonAntonio Jan 22 '14 at 20:50
@DonAntonio: I think he wants help evaluating the above sum. – Ron Gordon Jan 22 '14 at 20:53
Beats me, @RonGordon...and that equal sign to the left of the sum doesn't seem to clarify stuff: looks like something's missing. – DonAntonio Jan 22 '14 at 20:54
Thanks for the quick replies. There isn't an actual question, I'm just evaluating the time complexity of an algorithm, and one line requires that (the sum) amount of time to complete, but I need to show it in a form that isn't a summation. Like Gauss's summation of 1 -> i represented as n(n+1) / 2. – user2368124 Jan 22 '14 at 20:56

Using some well-known formulas, we get $$\sum_{i=1}^n \frac{i^2(i+1)}{2} = \frac{1}{2} \sum_{i=1}^n i^3 + \frac{1}{2} \sum_{i=1}^n i^2 = \frac{1}{2} \cdot \left( \frac{n(n+1)}{2} \right)^2 + \frac{1}{2} \cdot \frac{n(n+1)(2n+1)}{6}$$
Since evaluating the time complexity of an algorithm was mentioned, one can recall the following equivalent, valid for every $a\gt-1$: $$\sum_{i=1}^ni^a\sim\frac{n^{a+1}}{a+1}.$$ Using this for $a=3$ and $a=2$ yields $$\sum_{i=1}^n\frac{i^2(i+1)}2\sim\frac12\frac{n^4}{4}+\frac12\frac{n^3}3\sim\frac{n^4}8.$$ (Each sign $\sim$ above (read: equivalent to) has the quite precise meaning that the ratio of the RHS to the LHS converges to $1$ when $n\to\infty$.)