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It is well known that $$\sum_{r=1}^n r^\overline{m}=\frac {n^{\overline{m+1}}}{m+1}\\ \text{i.e.} \sum_{r=1}^n \scriptsize(r+1)(r+2)\cdots (r+m-1)=\frac{n(n+1)(n+2)\cdots (n+m)}{m+1}\\ \text{which can also be written as }\\ \qquad \scriptsize m!\sum_{r=1}^n\binom {r+m-1}m=\frac 1{m+1}\binom {n+m}{m+1}$$.

Can it be shown that $$\sum_{r=1}^n \left[r^\overline{m}\prod_{k=1}^q (r+p_k)\right] =\frac {n^{\overline{m+1}}}{m+1}P_q(n)\\ \text{i.e. }\tiny \sum_{r=1}^n r(r+1)(r+2)\cdots (r+m-1)(r+p_1)(r+p_2)\cdots (r+p_q) =\frac{n(n+1)(n+2)\cdots (n+m)}{m+1}\cdot \left(a_qn^q+a_{q-1}n^{q-1}+\cdots+a_1 q+a_0\right)$$ where $p_k$ are positive integers greater than $m$, and $P_q(n)$ is a polynomial of degree $q$ in $n$?

A few examples (with solutions from Wolframalpha):

1. $\scriptsize\displaystyle\sum_{r=1}^n \boxed{r(r+1)}(r+5)=\frac 1{4}\boxed{n(n+1)(n+2)}(n+7)$

2. $\scriptsize\displaystyle\sum_{r=1}^n \boxed{r(r+1)(r+2)}(r+4)(r+8)=\frac 1{12}\boxed{n(n+1)(n+2)(n+3)}(2n^2+30n+103)$

3.$\scriptsize\displaystyle\sum_{r=1}^n \boxed{r(r+1)(r+2)(r+3)}(r+5)(r+8)=\frac 1{105}\boxed{n(n+1)(n+2)(n+3)(n+4)}(15n^2+235n+884)$

Hypergeometricx
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1 Answers1

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The assumption is correct (and in fact can be generalized) since we have \begin{align*} \sum_{r=1}^nr^{\overline{m+1}}&=\frac{n^{\overline{m+2}}}{m+2}\\ &=\frac{n^{\overline{m+1}}}{m+1}\cdot\frac{m+1}{m+2}(n+m+1)\\ &=\frac{m+1}{m+2}(n+m+1)\sum_{r=1}^n r^{\overline{m}}\tag{1} \end{align*} as well as \begin{align*} \sum_{r=1}^nr^{\overline{m}} r&=\sum_{r=1}^nr^{\overline{m}}(r+m-m)\\ &=\sum_{r=1}^nr^{\overline{m+1}}-m\sum_{r=1}^nr^{\overline{m}}\tag{2}\\ \end{align*} and for positive integer $k$ \begin{align*} \sum_{r=1}^nr^{\overline{m}} r^k&=\sum_{r=1}^nr^{\overline{m}}(r+m-m)r^{k-1}\\ &=\sum_{r=1}^nr^{\overline{m+1}}r^{k-1}-m\sum_{r=1}^nr^{\overline{m}}r^{k-1}\\ \end{align*}

Let's look at OP's first example with \begin{align*} \sum_{r=1}^nr(r+1)&=\sum_{r=1}^nr^{\overline{2}}=\frac{1}{3}n(n+1)(n+2)\tag{3} \end{align*} We obtain \begin{align*} \color{blue}{\sum_{r=1}^nr^{\overline{2}}(r+5)} &=\sum_{r=1}^nr^\overline{2}r+5\sum_{r=1}^nr^{\overline{2}}\\ &=\left(\sum_{r=1}^nr^{\overline{3}}-2\sum_{r=1}^nr^{\overline{2}}\right)+5\sum_{r=1}^nr^{\overline{2}}\tag{apply 2}\\ &=\frac{3}{4}(n+3)\sum_{r=1}^nr^{\overline{2}}+3\sum_{r=1}^nr^{\overline{2}}\tag{apply 1}\\ &=\frac{3}{4}(n+7)\sum_{r=1}^nr^{\overline{2}}\\ &\,\,\color{blue}{=\frac{1}{4}n(n+1)(n+2)(n+7)}\tag{apply 3} \end{align*}

Markus Scheuer
  • 108,315