We have the sum
$$\sum_{i = 1}^n {i (i^m - (i - 1)^m)}$$
where $m$ and $n$ are positive integers. WolframAlpha will evaluate the sum for specific values of $m$, but returns no result when $m$ is left as a variable. I've tried to find a pattern in the results for specific values of $m$, with limited success. The first few expressions are:
- $\frac{1}{ 2} n (n + 1)$
- $\frac{1}{ 6} n (n + 1) (4 n - 1)$
- $\frac{1}{ 4} n^2 (n + 1) (3 n - 1)$
- $\frac{1}{30} n (n + 1) (24 n^3 - 9 n^2 - n + 1)$
- $\frac{1}{12} n^2 (n + 1) (10 n^3 - 4 n^2 - n + 1)$
- $\frac{1}{42} n (n + 1) (36 n^5 - 15 n^4 - 6 n^3 + 6 n^2 + n - 1)$
- $\frac{1}{24} n^2 (n + 1) (21 n^5 - 9 n^4 - 5 n^3 + 5 n^2 + 2 n - 2)$
- $\frac{1}{90} n (n + 1) (80 n^7 - 35 n^6 - 25 n^5 + 25 n^4 + 17 n^3 - 17 n^2 - 3 n + 3)$
- $\frac{1}{20} n^2 (n + 1) (18 n^7 - 8 n^6 - 7 n^5 + 7 n^4 + 7 n^3 - 7 n^2 - 3 n + 3)$
Except for the $m = 1$ case, which stands on its own, the results seem to come in pairs. The even-$m$ case is in the form [reciprocal of a number] $× n × (n + 1) ×$ [polynomial of degree $m - 1$]. The odd-$m$ case immediately following that is in the form [reciprocal of a smaller number] $× n^2 × (n + 1) ×$ [polynomial of same degree but with generally smaller or equal coefficients]. Beyond that, I'm stumped.
Is there a closed-form expression for this sum in terms of $m$ and $n$?
We have $$\begin{align*}\sum_{i=1}^ni(i^m-(i-1)^m)&=\sum_{i=1}^ni^{m+1}-\sum_{i=1}^n(i-1)^{m+1}+\sum_{i=1}^n(i-1)^m\\ &=n^{m+1}+\sum_{i=0}^{n-1}i^m.\end{align*}$$ Now, there is a well-known formula for $\sum_{i=0}^{n-1}i^m$ in terms of Bernoulli numbers.