I see this equation $$ 1^r + 2^r +... + n^r = a_1 n^1 + a_2 n^2 +... + a_{r+1} n^{r+1} $$ in Introduction to Linear Algebra, where $a_k$s are some constants.
How can I prove it?
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what?${}{}{}{}{}$ – Asinomás Dec 28 '16 at 01:59
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I wanna know how to prove this equation. can you help me? – Maybe Dec 28 '16 at 02:01
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what are those $a_k$'s? – Asinomás Dec 28 '16 at 02:01
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1See for example https://gowers.wordpress.com/2014/11/04/sums-of-kth-powers/ It has been asked before here, but can't find it right now – Winther Dec 28 '16 at 02:04
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don't you need an $a_0$ also? – Asinomás Dec 28 '16 at 02:04
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Where did you see this expression? In what book? – Dec 28 '16 at 02:06
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1Related: why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$ – Winther Dec 28 '16 at 02:11
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1@JorgeFernándezHidalgo Oh man, I love that post! – Simply Beautiful Art Dec 28 '16 at 02:17
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$s^r$ can be expressed as a linear combination of ${s \choose 0},{s \choose 1},...{s \choose r}$ where the constant on ${s \choose r}$ must be non zero. Then apply the hockey stick identity to show that the sum in question is a linear combination of ${n+1 \choose 1}, {n+1 \choose 2},....{n+1 \choose r+1}$, and again the constant on ${n+1 \choose r+1}$ is nonzero. Thus showing that sum is a polynomial of degree $r+1$.
Ahmed S. Attaalla
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Oh ok, I think I understand the question.
You want to prove that for a fixed $r$, there exists a degree $r+1$ polynomial $P$ with constant term $0$ such that $1^r+2^r+\dots+ n^r=P(n)$ for all $n\in\mathbb Z^+$.
This has been asked several times before, my favorite solution is with combinatorics and stirling coefficients, but you can find a bunch of solutions here
