I am wondering if there is a recursive formula to calculate $$S=1^{k}+2^{k}+3^{k}+\dots+n^{k}$$ Where $k$ and $n$ are natural numbers.
Asked
Active
Viewed 113 times
2
-
1$f(x) = f(x-1)+x^k?$ – Fallen Jul 08 '13 at 14:09
-
I intended to find something depending on k in the recurrence,not the n.... – Denis Jul 08 '13 at 14:12
-
I was looking for this clarification indeed. That's why didn't add it as answer :) – Fallen Jul 08 '13 at 14:13
-
Googling Faulhaber's formula should get you something decent. – Ragib Zaman Jul 08 '13 at 14:17
-
1Check this paper. – user5402 Jul 08 '13 at 14:25
-
You can generalize this technique. – Mhenni Benghorbal Jul 08 '13 at 14:43
1 Answers
0
I believe this should do,
$$f[k][n]=f[k][n-1]+(f[k-1][n]-f[k-1][n-1])\cdot n$$
where $f[k][n] = 1^k+2^k+\dots+n^k$. so we get $n^{k-1} = (f[k-1][n]-f[k-1][n-1])$ and multiplying both side with $n$ we get $n^k=(f[k-1][n]-f[k-1][n-1])\cdot n$.
And $f[k][n-1]+n^k = f[k][n]$
Fallen
- 236