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I understand the process in calculating a simple series like this one

$$\sum_{n=0}^4 2n$$

but I do not understand the steps to calculate a series like this one

$$\sum_{n=1}^x n^2$$

I have an awesome calculator and know how to use it, so I know the solution

$$\frac{x^3}{3} + \frac{x^2}{2} + \frac{x}{6}$$

but I don't know the steps. All the explanations I find (like those on Purple Math) are for summations like the first one I listed. If someone can provide a detailed explanation on how to calculate a summation like mine or provide a link to one it would be much appreciated.

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I think you can use differences to find such formulas. I don't know the details though since I only saw it one in a class a long time ago and the details were not presented, just some examples of how to do it. –  Adrián Barquero Feb 27 '11 at 3:19
    
I'll try to find a reference for that method. –  Adrián Barquero Feb 27 '11 at 3:22
    
Here's a reference: stanford.edu/~dgleich/publications/finite-calculus.pdf. On page 16 of this pdf an actual derivation of the formula is shown. This is a process that can be generalized to many summations. Also, a Google search for "finite calculus" or "calculus of finite differences" can be very prolific. –  Abel Feb 27 '11 at 6:09

3 Answers 3

up vote 5 down vote accepted

Yes, you can use differences to find such formulas. Many times there are also easier ways (involving tricks). For this one you can use that

$$\sum_{i=0}^n i^3 = \left ( \sum_{i=0}^n (i+1)^3 \right ) - (n+1)^3$$

From here it is an easy computation to find the answer:

$$\sum_{i=0}^n i^3 = \sum_{i=0}^n i^3 + \sum_{i=0}^n 3i^2 + \sum_{i=0}^n 3i + \sum_{i=0}^n 1 -(n+1)^3$$

which gives

$$\sum_{i=0}^n i^2 = \frac{-\frac{3n(n+1)}{2} - n -1 + n^3 +3n^2 +3n +1}{3} $$

$$\sum_{i=0}^n i^2 = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}$$

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Excellent, I actually remember seeing that trick now. I went back to my notes and there it was! Thanks for your help. –  ubiquibacon Feb 27 '11 at 3:58
    
as I look over your answer again, I want to make sure, did you forget to make i=1 in the last part? Also (since my calculator tells me different) I want to make sure that in the first line you are making i=0 because you are now raising i to the power of 3 (i^3) instead of the power of 2 (i^2)... is that correct? –  ubiquibacon Feb 27 '11 at 4:17
    
@typoknig In the end it doesn't matter since $0^2 = 0$ will not change the sum - you may replace it with a $1$ if you wish. However, $i=0$ is needed in the first equality: you would otherwise have $\sum_{i=1}^n (i+1)^3 = 2^3 + 3^3 + ... + (n+1)^3$, without the $1$. In fact, anywhere from then on you may index from $1$. –  milcak Feb 27 '11 at 4:22
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I would like to add: this method can be applied to compute a closed-form expression for $\sum_{i=1}^n i^k$ knowing one for $\sum_{i=1}^n i^{k-1}$, and is presented in Spivak's Calculus. –  Abel Feb 27 '11 at 6:13

$$(k+1)^3-k^3 = 3k^2 + 3k + 1$$ Now sum it up from $k=1$ to $n$. Notice the telescopic summation on the left side and use $\displaystyle \sum_{k=1}^n k = \frac{n(n+1)}{2}$ to get the answer. This is a standard technique to compute such sums.

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Oh I see, and that equation on the right actually looks familiar... the gauss formula right? –  ubiquibacon Feb 27 '11 at 3:30
    
@typoknig : Yes –  user17762 Feb 27 '11 at 3:32
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I like this more than the choosen answer because it is easily seen to be true. –  Apprentice Queue Feb 27 '11 at 6:52
    
This was good info for me, but knowing the little trick in the answer I selected will likely make my life easier in the very near future. –  ubiquibacon Feb 27 '11 at 7:20
    
Actually, what you call a trick is the same as the "telescopic differences" approach described here. –  Wok Feb 27 '11 at 9:18

Another approach is to know (guess) that $\sum_{n=1}^xn^2$ is cubic in $x$. Similar to an integral, the sum adds one degree. Then you can just say $\sum_{n=1}^xn^2=Ax^3+Bx^2+Cx+D$. If you calculate that the sum up to 1,2,3,4 is 1,5,14,30 (or start with 0 and sum to 0 makes it a bit easier) you can just solve the simultaneous equations for A,B,C, and D.

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I guess my problem is seeing the relation between the summations in their closed form and their factored equivalent. I understand well enough now to muddle my way through, and many sites show the factored equivalents of summations, but I would still like to see how I could factor a closed form summation with a higher order n (like maybe n^7, or something higher) that hasn't already been pre-calculated for me on some site. Finding how to manually calculate the factored equivalent of closed form summations was my main goal with this question, but I didn't know that when I asked :) –  ubiquibacon Feb 27 '11 at 8:42

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