$$\sum\limits_{i=1}^{n}i^x = P_{x+1}(n)$$ Let x be any nonnegative integer and show that there is a polynomial $P_{x+1}$ of degree $x+1$ for every $n$ greater than or equal to $1$.
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why using induction? – ShakesBeer Oct 09 '14 at 23:13
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it doesn't have to be induction if you can do it another way. – user181415 Oct 09 '14 at 23:15
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See Faulhaber's formula. – Lucian Oct 10 '14 at 00:10
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wow it's faulhaber's formula – user181415 Oct 10 '14 at 00:21
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does it also give the proof on the faulhaber's formula wikipedia page? – user181415 Oct 10 '14 at 00:32
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Hint Prove that there exists a polynomial $Q_{x+1}(n)$ with the property that [ Q_{x+1}(n+1)-Q_{x+1}(n)=n^x \,. ]
You can prove this for example by observing that $n+1-n, (n+1)^2-n^2,.., (n+1)^{x+1}-n^{x+1}$ are linearly independent, thus a basis for the space of polynomials of degree at most $x$ in the variable $n$.
Once you do this, you can easily prove that for the right constant $C$, which can be obtained by setting $n=1$, the polynomial [ P_{x+1}(n)=Q_{x+1}(n)+C \,, ] works.
Second solution
$$(i+1)^{x+1}=i^{x+1}+(x+1)i^x +\sum_{j=2}^{x+1} \binom{x+1}{j} i^{k+1-j} \,.$$
Sum this from $i=1$ to $n$, cancel the common terms, and use induction by $x$.
N. S.
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so do I not need to do this by induction? I can prove this the way you have said? – user181415 Oct 09 '14 at 23:44
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N.S. can you like elaborate more or do something that helps me understand this better? – user181415 Oct 09 '14 at 23:57