Proving Polynomial Expression: 1^k+2^k+...+n^k as a Degree-k+1 Polynomial

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The discussion centers on proving that the sum of powers, 1^k + 2^k + ... + n^k, can be expressed as a polynomial of degree k+1 in n. Some participants argue that the proof is trivial, suggesting that it can be represented in various forms, including polynomial expressions. However, others emphasize that for it to be a valid polynomial, the coefficients must be constant, which complicates the assertion. The conversation highlights the need for a rigorous proof, likely through induction, to establish the polynomial nature of the expression for all n. Ultimately, the consensus is that without a proper solution, the problem remains unresolved.
Icebreaker
"Prove that

1^k+2^k+...+n^k

can be written as a polynomial in n of degree at most k+1."

Isn't this kinda trivial? I mean I know the "book" solution is to prove by induction, etc, but assuming that I have the above expression, I can prove or disprove it, depending on how I interpret the question.

If it means that it can be written in the above conditions AND NOTHING ELSE, I can easily produce a counterexample:

1+2+3 = 3^3 - 7\times3

If it means that it can be written in the above conditions, but does not prohibit the existence of other solutions, then it's trivial, because the above expression can be written as

an^{k+1} for some real number a
 
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Icebreaker said:
If it means that it can be written in the above conditions AND NOTHING ELSE, I can easily produce a counterexample:

1+2+3 = 3^3 - 7\times3

It doesn't mean "and nothing else", it's already written in a form that you wouldn't call a polynomial.

Icebreaker said:
If it means that it can be written in the above conditions, but does not prohibit the existence of other solutions, then it's trivial, because the above expression can be written as

an^{k+1} for some real number a

No choice of a here will hold for all n. Your polynomial is supposed to equal that expression for all values of n.
 
True, no choice of a will hold for every n, but there exists one for every n.
 
Icebreaker said:
True, no choice of a will hold for every n, but there exists one for every n.

It's not a polynomial if the coefficients aren't constant. Your a will depend on n in some unspecified way, and you haven't solved the problem.
 
No easy way out then. Damn.
 
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