general way to get formula as multiplication.

Question

We assume: $$ 1^n + 2^n + 3^n + .. + k^n$$ where k and n are natural numbers. Are there a general way to get it as multiplication?

For example: $$ 1^3 + 2^3 + 3^3 + .. + k^3 = \binom {k+1} 2 ^2 $$

Answer 1

You can compute it recursively. Let us introduce a notation: $S_{p,I} := \sum_{i=0}^I i^p$

You know that $(1+r)^n-r^n=\sum_{(k=0)}^{n-1} \binom {n}{k} r^k$

Thus $$(R+1)^n = \sum_{r=0}^{R} \sum_{(k=0)}^{n-1} \binom {n}{k} r^k = \sum_{(k=0)}^{n-1} \binom {n}{k} \sum_{r=0}^{R} r^k = \sum_{(k=0)}^{n-1} \binom {n}{k} S_{k,R} $$

Answer 2

Suppose that for $0\le j\lt m$, there is a polynomial $P_j$, with degree $j+1$, so that $$ \sum_{k=0}^nk^j=P_j(n)\tag{1} $$ The binomial theorem says that $$ (k+1)^{m+1}-k^{m+1}=\sum_{j=0}^m\binom{m+1}{j}k^j\tag{2} $$ Sum $(2)$ in $k$ using $(1)$: $$ \begin{align} (n+1)^{m+1} &=\sum_{k=0}^n\sum_{j=0}^m\binom{m+1}{j}k^j\\ &=\sum_{k=0}^n\left[(m+1)k^m+\sum_{j=0}^{m-1}\binom{m+1}{j}k^j\right]\\ &=(m+1)\sum_{k=0}^nk^m+\sum_{j=0}^{m-1}\binom{m+1}{j}P_j(n)\tag{3} \end{align} $$ Solving $(3)$ for $\sum\limits_{k=0}^nk^m$ $$ \begin{align} \sum_{k=0}^nk^m &=\frac1{m+1}\left[(n+1)^{m+1}-\sum_{j=0}^{m-1}\binom{m+1}{j}P_j(n)\right]\\ &=P_m(n)\tag{4} \end{align} $$ That is, $(1)$ is true for $j=m$.

Thus, we have shown that if $(1)$ is true for $0\le j\lt m$, then $(1)$ is also true for $j=m$. Since $P_0(n)=n+1$, we have that $(1)$ is true for $j=0$. Therefore, by induction, $(1)$ is true for all $j\ge0$.

Notice that $(4)$ also gives us a formula to compute $P_m$ from the $P_j$ for $j\lt m$. $$ \begin{align} P_1(n) &=\frac12\left[\vphantom{\binom{2}{0}}\right.(n+1)^2-\binom{2}{0}\overbrace{(n+1)\vphantom{\frac{n^2+n}2}}^{P_0(n)}\left.\vphantom{\binom{2}{0}}\right]\\[4pt] &=\frac{n^2+n}2\\ P_2(n) &=\frac13\left[\vphantom{\binom{2}{0}}\right.(n+1)^3-\binom{3}{0}\overbrace{(n+1)\vphantom{\frac{n^2+n}2}}^{P_0(n)}-\binom{3}{1}\overbrace{\frac{n^2+n}2}^{P_1(n)}\left.\vphantom{\binom{2}{0}}\right]\\[4pt] &=\frac{2n^3+3n^2+n}6\\ P_3(n) &=\frac14\left[\vphantom{\binom{2}{0}}\right.(n+1)^4-\binom{4}{0}\overbrace{(n+1)\vphantom{\frac{n^2+n}2}}^{P_0(n)}-\binom{4}{1}\overbrace{\frac{n^2+n}2}^{P_1(n)}-\binom{4}{2}\overbrace{\frac{2n^3+3n^2+n}6}^{P_2(n)}\left.\vphantom{\binom{2}{0}}\right]\\[4pt] &=\frac{n^4+2n^3+n^2}4 \end{align} $$