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/math/ - Mathematics


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25 Dec 2021Mathchan is launched into public

3 / 1 / 3 / ?

File: 25d30fdff6e5edb9cd299f196780830a.png ( 216.3 KB , 1200x899 , 1662422662595.png )

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The sum of the coefficients of the expanded Sum of the n first postive integer powers seem to equal the denominator.
Seems pretty cool and I can't find anything about it online
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Well this can be fairly easily explained by noting that the sum of the coefficients of a polynomial equals the polynomial evaluated at 1, as well the LHS for n=1 always 1. (btw it should be the sum from k=1 to n of k^something, not vice versa, your picture includes this typo consistently)
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Is there a formula for the general case, e.g. n^a?
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>>469
Yep! sum_{n=0}^k n^a=
=(Bernoullipolynomial_(a+1)(k+1)-Bernoullinumber_(a+1))/(a+1)=
=harmonicnumber(k,-a)=
=-Hurwitzzeta(-a,k+1)+Riemannzeta(-a) for a in Z+