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1^n + 2^n + 3^n ... a^n

?

- Thread starter Werg22
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- #1

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1^n + 2^n + 3^n ... a^n

?

- #2

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well, as far as i know, the formulas differ depending on the power. I don't think that you can generalize a rhiemann summation like that...although it would defenetly be useful.Werg22 said:

1^n + 2^n + 3^n ... a^n

?

i know that sum(n) = n(n-1)/2

sum(n^2)=n(n-1)(n-2)/6

and so on...but i never had to memorize them so i might be wrong about the second one...if there was a way to combine all of them no matter the power, it should be in the precalculus manuals i think.

- #3

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I was looking for this a few weeks ago and found one here:

http://mathworld.wolfram.com/FaulhabersFormula.html" [Broken]

Also, google: "general power series sum" or variation and you should get quite a few hits

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- #4

mathman

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I came up with another general way. Borrows a bit from Faulhaber's formula but it's a little simpler and might be a little faster as it uses fewer factorials and no Bernoulli numbers

[tex]

\sum_{j=0}^m j^p \quad = \quad p! \sum_{j=0}^m Q_{pj} R_{pmj}

[/tex]

[tex]

Q_{pk} \quad = \quad \left\{

\begin{array}{ll}

\sum_{j=0}^k -1^j \frac{ (k - j + 1)^p}{j!(p - j)!} \quad & k < p \\

1 & k \geq p

\end{array}

[/tex]

[tex]

R_{pmk} \quad = \quad \prod_{l=1}^p 1 + \frac{m - l}{k}

[/tex]

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- #6

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There's a typo there, it should beRobokapp said:well, as far as i know, the formulas differ depending on the power. I don't think that you can generalize a rhiemann summation like that...although it would defenetly be useful.

i know that sum(n) = n(n-1)/2

sum(n^2)=n(n-1)(n-2)/6

and so on...but i never had to memorize them so i might be wrong about the second one...if there was a way to combine all of them no matter the power, it should be in the precalculus manuals i think.

sum(n^2)=n(n+1)(2n+1)/6.

Although n(n-1)(n-2)/6 will always return a possitve integer for n > 2, it does not give you the desired number.

- #7

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Here's one using Lagrange Interpolating Polynomials (eqivalent to simplified polynomial curve fitting here).

[tex]\sum_{k=1}^{n}k^{p}=\sum_{k=1}^{p+2}\left(\sum_{m=1}^{k}m^{p}\prod_{j=1}^{p+2}\frac{n-j}{k-j}\right)[/tex]

where, in the product, the terms with j=k are omitted.

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- #9

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You can use bivariate induction to prove the above sum.Werg22 said:

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