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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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    Here's a reference: http://www.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.2011-02-27

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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 :)2011-02-27
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$(k+1)^3-k^3 = 3k^2 + 3k + 1$ 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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    Actually, what you call a trick is the same as the "telescopic differences" approach described here.2011-02-27
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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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    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.2011-02-27