Friend of mine, have found a formula for the summation $\sum \limits_{i=0}^{n-1} (a + d * i )^k$ for given $k,n \in \mathbb{N}$ and $a,d \in \mathbb{R}$.
He checked it for many values, and professor in BGU proved it.
The formula he found is recursive so letting $$p(a,d,k,n) = \sum \limits_{i=0}^{n-1} (a + d * i )^k$$ the recursive step is on $k$ and he is asking if there is a recursive formula in this form known to the math commuinty ??
In this form : $$p(a,d,k,n) = f(a,d,k,n)+\sum \limits_{i=0}^{n-1}p(a,d,k-i,n) g(a,d,k,n,i)$$
Where $f,g$ are simple function consists of factorials and powers only(which are much simpler than Bernoulli numbers or Stirling numbers).
Any ref to existing formula would be more than helpful.
Thanks.
We have that $$ \eqalign{ & p(a,d,k,n) = \sum\limits_{i = 0}^{n - 1} {(a + di)^{\,k} } = \sum\limits_{i = 0}^{n - 1} {(a + di)(a + di)^{\,k - 1} } = \cr & = a\sum\limits_{i = 0}^{n - 1} {(a + di)^{\,k - 1} } + d\sum\limits_{i = 0}^{n - 1} {i\;(a + di)^{\,k - 1} } \cr} $$ and applying the Summation by parts to the second sum, we get $$ \eqalign{ & \sum\limits_{i = 0}^{n - 1} {i\;(a + di)^{\,k - 1} } = \,\left( {n - 1} \right)p(a,d,k - 1,n) + p(a,d,k - 1,0) - \sum\limits_{i = 0}^{n - 1} {p(a,d,k - 1,i)} = \cr & = \,\left( {n - 1} \right)p(a,d,k - 1,n) - \sum\limits_{i = 0}^{n - 1} {p(a,d,k - 1,i)} \cr} $$ i.e. $$ p(a,d,k,n) = \left( {a + d\left( {n - 1} \right)} \right)\,p(a,d,k - 1,n) - d\sum\limits_{i = 0}^{n - 1} {p(a,d,k - 1,i)} $$
Iterating that $k$ times, to reach to $p(a,d,0,n)$, will provide a linear recursion in $p(a,d,k-j,n)$, with coefficients that are polynomials in $a,d,n$.
So, sorry to tell that , although not knowing the details, there should be nothing special in your friend's result.
Yet, it is much appreciable if he is fresh in this field.