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For any positive integer $n$ and $m,$ I was wondering if there is any way to get a closed formula for $$S(n,m)=1^m+2^m+3^m+\cdots+n^m$$ something like $$S(n,1)=1+2+3+\cdots+n=\frac{n(n+1)}{2}.$$

mathguy
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  • Look for Stirling numbers of the second kind (http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html) – Jack D'Aurizio Feb 27 '15 at 18:58
  • Yes this is known as Faulhaber's formula. Check out the Wikipedia article. – Tim Raczkowski Feb 27 '15 at 18:59
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    Just in Wikipedia http://en.wikipedia.org/wiki/Faulhaber%27s_formula – Elaqqad Feb 27 '15 at 18:59
  • Have a look at this http://math.stackexchange.com/questions/1160658/how-to-find-general-formula-of-sum-n2/1160711#1160711 It gives you a general approach to find the above summation – Fermat Feb 27 '15 at 19:23
  • This question is the source of one of the most condescending mathematical quotations in history. Jacob Bernoulli, after making a table up to m=10 and describing how to continue it, says "From this it will become clear how useless was the work of Ismael Bullialdus spent on the compilation of his voluminous Arithmetica Infinitorum in which he did nothing more than compute with immense labour the sums of the first six powers, which is only a part of what we have accomplished in the space of a single page." – Barry Smith Feb 27 '15 at 19:30
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    possible duplicate of Is there a general product formula for $\sum\limits_{k=1}^{n} k^p$ –  Feb 27 '15 at 20:04

3 Answers3

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This answer is the exact copy of Garvil's answer and so I've made this answer 'community wiki'.

Consider any natural number $r$. You have $$r^3-(r-1)^3=3r^2-3r+1.$$

Now telescope it: $$ 1^3-0^3=3-3+1 $$ $$2^3-1^3=3\cdot2^2-3\cdot2+1 $$ $$\vdots $$ $$ n^3-(n-1)^3=3n^2-3n+1 $$ Now add, and see them cancel out. You are left with $$n^3=3(1^2+2^2+\cdots+ n^2)-3(1+2+3+\cdots+n)+n$$ You must know $$ 1+2+3+\cdots+n=\frac{n(n+1)}{2}. $$ Plug it in, and you get the answer. Also, please see that this method works even for $\sum r^4,r^5,\cdots$. I have tried it out. All you need is the sum of its previous powers.

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user26486
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Read this: http://www.codeproject.com/Tips/792255/Faulhaber-made-easy.

Closed formulas are also known.

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[Copied from an answer of mine at https://math.stackexchange.com/questions/1160658/how-to-find-general-formula-of-sum-n2/1160711#1160711]

There is an interesting method to find the above summations.

Theorem. Let $n\geq r$. Then $$\sum_{k=1}^{n}{k\choose r}={r\choose r}+...+{n\choose r}={k+1\choose r+1}$$ note that for $k<r$, $c(k,r)=0$.

Now, since $k^2=k+k(k-1)={k\choose 1}+2{k\choose 2}$ taking sum of both sides and using the above theorem gives $$\sum_{k=1}^{n}k^2=\sum_{k=1}^{n}{k\choose 1}+2\sum_{k=1}^{n}{k\choose 2}={n+1\choose 2}+2{n+1\choose 3}=\frac{n(n+1)(2n+1)}{6}$$ I said this way of proof is more interesting because for the summation $\sum_{k=1}^{n}k^3$ one can write $$k^3=x{k\choose 1}+y{k\choose 2}+z{k\choose 3}$$ and then finds the coefficients $x,y,z$ by solving a system and finally by applying the above theorem gets $\sum_{k=1}^{n}k^3=\left(\frac{n(n+1)}{2}\right)^2$.

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Fermat
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