# Is there a closed-form expression for this sum in terms of $m$ and $n$?

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:

1. $$\frac{1}{ 2} n (n + 1)$$
2. $$\frac{1}{ 6} n (n + 1) (4 n - 1)$$
3. $$\frac{1}{ 4} n^2 (n + 1) (3 n - 1)$$
4. $$\frac{1}{30} n (n + 1) (24 n^3 - 9 n^2 - n + 1)$$
5. $$\frac{1}{12} n^2 (n + 1) (10 n^3 - 4 n^2 - n + 1)$$
6. $$\frac{1}{42} n (n + 1) (36 n^5 - 15 n^4 - 6 n^3 + 6 n^2 + n - 1)$$
7. $$\frac{1}{24} n^2 (n + 1) (21 n^5 - 9 n^4 - 5 n^3 + 5 n^2 + 2 n - 2)$$
8. $$\frac{1}{90} n (n + 1) (80 n^7 - 35 n^6 - 25 n^5 + 25 n^4 + 17 n^3 - 17 n^2 - 3 n + 3)$$
9. $$\frac{1}{20} n^2 (n + 1) (18 n^7 - 8 n^6 - 7 n^5 + 7 n^4 + 7 n^3 - 7 n^2 - 3 n + 3)$$

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.
• It also looks like Faulhaber's formula is for $\sum_{i = 1}^n i^m$, not $\sum_{i = 0}^{n -1} i^m$. Jul 5 at 15:28
• @Lawton - $i = (i-1) +1$, but there is a sign error. The final sum should also be subtracted. As for Faulhaber's formula, just add $0$ to it to cover the $i = 0$ term (when $m \ne 0$), and apply the formula with $n$ replaced by $n-1$. Jul 5 at 17:23