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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 $n$?

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1 Answer 1

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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.

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  • $\begingroup$ I'm not seeing how the first expression in your answer is true. It seems like the right side is not equal to the left side. Am I missing something? $\endgroup$
    – Lawton
    Jul 5 at 15:26
  • $\begingroup$ It also looks like Faulhaber's formula is for $\sum_{i = 1}^n i^m$, not $\sum_{i = 0}^{n -1} i^m$. $\endgroup$
    – Lawton
    Jul 5 at 15:28
  • $\begingroup$ @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 $n-1$. $\endgroup$ Jul 5 at 17:23

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