prove that $\binom{k+1}{1} S_k(n)+\binom{k+1}{2} S_{k-1}(n)+.........+\binom{k+1}{k+1} S_0(n)=(n+1)^{k+1}-1$

Question

For $n,k \in N$,we define

$S_k(n)=1^k +2^k+... +n^k$

if $(1+x)^p=1+\binom{p}{1}x+\binom{p}{2}x^2+.... +\binom{p}{p}x^p,p\in N$,then we have to prove that

$\binom{k+1}{1} S_k(n)+\binom{k+1}{2} S_{k-1}(n)+...+\binom{k+1}{k+1} S_0(n)=(n+1)^{k+1}-1$

this looks similar to $\binom{k+1}{1}n+\binom{k+1}{2}n^2+...+\binom{k+1}{k+1}n^{k+1}= (1+n)^{k+1}-\binom{k+1}{0}=(1+n)^{k+1}-1$

what next?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{\ell = 1}^{k + 1}{k + 1 \choose \ell}\sum_{m = 1}^{n}m^{k -\ell + 1} & = \sum_{m = 1}^{n}m^{k + 1}\sum_{\ell = 1}^{k + 1}{k + 1 \choose \ell} \pars{1 \over m}^{\ell} \\[5mm] & = \sum_{m = 1}^{n}m^{k + 1}\bracks{\pars{1 + {1 \over m}}^{k + 1} - 1} \\[5mm] & = \sum_{m = 1}^{n}\pars{m + 1}^{k + 1} - \sum_{m = 1}^{n}m^{k + 1} = \sum_{m = 2}^{n + 1}m^{k + 1} - \sum_{m = 1}^{n}m^{k + 1} \\[5mm] & = \bracks{-1 + \sum_{m = 1}^{n}m^{k + 1} + \pars{n + 1}^{k + 1}} - \sum_{m = 1}^{n}m^{k + 1} \\[5mm] & = \bbx{\pars{n + 1}^{k + 1} - 1} \end{align}

Answer 2

By binomial theorem:

$$(1+x)^{m+1}-x^m=1+{n \choose 1}(x)+{n \choose 2} (x^2)+....+{m+1 \choose m} (x^m)$$ putting $x=1,2,3,,..n$ and adding them, we get $$(n+1)^{n+1}-1=(1+1+1...+1)+{m+1 \choose 1}(1+2+3+...+n)+{m+1 \choose 2}(1^2+2^2+3^2++...+n^2)+{m+1 \choose 3} (1^3+2^3+3^3+4^3+...+n^3)+.....+{m+1 \choose m}(1^m+2^m+3^m+4^m+...+n^m)$$ $$\implies \sum_{k=0}^{m} S_k {m+1 \choose k}=(n+1)^{m+1}-1,~~ S_k=1^k+2^k+3^k+4^k+...n^k$$