On sums of powers of natural numbersThe problem of finding the sum of a polynomial's values is considered. In particular, for any $n\geq 3$, the explicit formula for the sum of the $n$th powers of natural numbers $S_n=\sum_{x=1}^{m}x^{n}$ is proved: $$\sum_{x=1}^{m}x^{n}=(-1)^{n}m(m+1)(-\frac{1}{2}+\sum_{i=2}^{n}a_i(m+2)(m+3)...(m+i)),$$ here $a_i=\frac{1}{i+1}\sum_{k=1}^{i}\frac{(-1)^{k}k^{n}}{k!(i-k)!}$, $(i=2,3,...,n-1)$, $a_n=\frac{(-1)^n}{n+1}$. Note that this formula does not contain Bernoulli numbers.
arXiv.org 
