Is there a general algorithm to generate the coefficients of the expansion of ${\displaystyle f_{k}(n)=\sum _{i=1}^{n}i^{k}}$? For example, ${\displaystyle f_{7}(n)={\frac {n^{8}}{8}}+{\frac {n^{7}}{2}}+{\frac {7n^{6}}{12}}-{\frac {7n^{4}}{24}}+{\frac {n^{2}}{12}}}$. I could use polynomial interpolation on the first ${\displaystyle k+2}$terms of the series, but that gets impractical quickly if ${\displaystyle k}$is large. 24.255.17.182 (talk) 21:47, 25 November 2016 (UTC)[reply]

One interesting thing I noticed is that ${\displaystyle {\frac {n^{k-i}(i+1)!}{\operatorname {C} (k,i-1)}}}$is invariant with respect to ${\displaystyle k}$, and starts out ${\displaystyle 1,{\frac {1}{2}},0,-1,0,20,0,-1512,0,302400,0,-131345280,0,108972864000,0,157661939635200,0,371498581149696000...}$, but I don't see an obvious pattern here and this series isn't in OEIS. 24.255.17.182 (talk) 22:13, 25 November 2016 (UTC)[reply]

(ec)Use Binomial_coefficient#Binomial_coefficients_as_a_basis_for_the_space_of_polynomials and the Hockey-stick identity.

${\displaystyle \sum _{i=1}^{n}i^{k}=\sum _{i=1}^{n}\sum _{m=0}^{k}a_{m}{\binom {i}{m}}=\sum _{m=0}^{k}a_{m}\sum _{i=1}^{n}{\binom {i}{m}}=\sum _{m=0}^{k}a_{m}{\binom {n+1}{m+1}}}$

Bo Jacoby (talk) 22:24, 25 November 2016 (UTC).[reply]

Sorry if it's obvious but could you explain how ${\displaystyle a_{m}}$is to be computed? 24.255.17.182 (talk) 23:03, 25 November 2016 (UTC)[reply]

I think what you're looking for is Faulhaber's formula. A generalization is the Euler–Maclaurin formula. --RDBury (talk) 01:43, 26 November 2016 (UTC)[reply]

The formula is found in the link I gave you.

${\displaystyle a_{m}=\sum _{i=0}^{m}(-1)^{m-i}{\binom {m}{i}}i^{k}.}$

Bo Jacoby (talk) 07:18, 26 November 2016 (UTC).[reply]