How is the summation formula's for $x,x^2,x^3,x^4,\ldots$ derived? I know how to do it for $x$ which is $n^2/2 + n/2$ but I am having hard time deriving the summation formula for $x^n$ on my own. I heard the best way to do it was by binomial coefficient but I don't see it. Before posting, I read couple of posts but none helped. So if you know and direct me towards any post or a webpage (wiki) that can help, that would be great. If not, any input would be much appreciated.
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1See http://en.wikipedia.org/wiki/Sums_of_powers. – lhf Jun 04 '13 at 02:30
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Didn't we already have this question today? – Zen Jun 04 '13 at 02:33
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@Zen We did. ${}{}{}$ – Pedro Jun 04 '13 at 02:34
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1Where? Where? Where?.... ;) – Eames Cobb Jun 04 '13 at 02:35
5 Answers
I can show you by the example of $x^2$, and I hope that you can see how to generalize the method. Begin with: $$\displaystyle\sum\limits_{i=1}^n i^2 = \int_0^n\!x^2 dx\ + \zeta(n)$$ With this, a simple integral can be carried out, and the only work left to do is to determine $\zeta(n)$, the error resulting from using an integral approximation.
To do so, consider the figure shown above, a snippet of the $y=x^2$ curve. The shaded region represents the area not captured in the integral and thus is the error. Any particular area of shading, $A$, for example the area from $x=1$ to $x=2$, can be represented with this:
$$ A(a) = \int\limits_{a^2}^{(a+1)^2} \int\limits_{a}^{sqrt(y)}1\,dxdy = \int\limits_{a^2}^{(a+1)^2}sqrt(y) - a \,dy $$ And this simplifies nicely to: $$ A(a) = a + \frac{2}{3} $$ Now, we add these up to determine the total error: $$ \zeta(n)\ = \displaystyle\sum\limits_{a=0}^{n-1} a+\frac{2}{3}\ = \displaystyle\sum\limits_{a=1}^n a-1+ \frac{2n}{3} $$ Finally, $$\zeta(n)\ = \frac{n(n+1)}{2}\ - \frac{n}{3}$$ Then, the total summation results from adding this result to the first integral. $$\displaystyle\sum\limits_{i=1}^{n}i^2\ = \frac{n^3}{3} + \frac{n(n+1)}{2}\ - \frac{n}{3} = \frac{n(n+1)(2n+1)}{6} $$

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For the sum of $x^2$, I hope this The sum of the squares of the first n natural numbers could be helpful. Actually, you can follow the same method for $x^n$.

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On the linked page, I don't find the line under "We add this n identities and we get:" to be well justified. The last three terms on the right are well explained, but the term on the left and the first term on the right are not the sum of the corresponding terms above. You can cancel a lot of terms to get there, but that should be noted. – Ross Millikan Jun 04 '13 at 02:53
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For the general solution, the easiest path is to use the Euler-Maclaurin formula to compute the sum: $$ S_m(n) = \sum_{0 \le k \le n - 1} k^m $$ as: $$ S_m(n) = \frac{1}{m + 1} \, \left( \sum_{0 \le k \le m} \binom{m + 1}{k} \, B_k n^{m + 1 - k} \right) $$ Here $B_k$ are the Bernoulli numbers

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One way to do is is to define falling factorial powers as: $$ x^{\underline{m}} = x (x - 1) \ldots (x - m + 1) $$ It is easy to prove by induction that: $$ \sum_{1 \le k \le n} k^{\underline{m}} = \frac{n^{\underline{m + 1}}}{m + 1} $$ There is one way to express $x^m$ as a combination of falling factorial powers, given by Stirling numbers of the second kind: $$ x^m = \sum_{0 \le k \le m} \genfrac{\{}{\}}{0pt}{}{m}{k} x^{\underline{k}} $$ (or you can just set up the relevant equations, they are easy to solve by hand for any particular $m$). The result is then easy: $$ \sum_{1 \le k \le n} k^m = \sum_{1 \le r \le m} \frac{1}{r + 1} \genfrac{\{}{\}}{0pt}{}{m}{r} n^{\underline{r + 1}} $$ This is convenient for calculation for smallish $m$.

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Here's a brainless approach: When summing $x^k$, you expect a polynomial of order $x^{k+1}$. There are a number of fuzzy reasons why this is the case, e.g. because integration is similar to summing, cf. michael10000's answer.
In any case, a polynomial of order $n$ is determined by its values at $n + 1$ points, and it's easy to calculate the first $n + 1$ values of this polynomial, so you can just use any number of methods for polynomial interpolation (I wrote an article leading you through one method: Interpolating Polynomials) to work out what all the coefficients are.

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