Endliche Summen(1)

\begin{equation} \sum\limits_{k=1}^{n}{k} = \frac{n\left( {n+1} \right)}{2} \end{equation}

\begin{equation} \sum\limits_{k=1}^{n}{k^2} = \frac{n\left( {n+1} \right)\left( {2n+1} \right)}{6} \end{equation}

\begin{align} \sum\limits_{k=1}^{n}{k^3} &= {\left[ {\frac{n\left( {n+1} \right)}{2}} \right]}^{2} \\&= {\left[ {\sum\limits_{k=1}^{n}{k}} \right]}^{2}\\ &= \sum\limits_{i=1}^{n}{\sum\limits_{j=1}^{n}{ij}} \end{align}

\begin{equation} \sum\limits_{k=1}^{n}{k^4} =\frac{n\left( {n+1} \right)\left( {2n+1} \right)\left( {3n\left( {n+1} \right)-1} \right)}{30} \end{equation}

\begin{equation} \sum\limits_{k=1}^{n}{k^5} =\frac{{n^2}{\left( {n+1} \right)}^2 {\left( {2n\left( {n+1} \right)-1} \right)}}{12}\end{equation}

\begin{equation} \sum\limits_{k=1}^{n}{k^6} =\frac{n\left( {n+1} \right)\,\left( {2n+1} \right)\,\left[ {3n\left( {n+1} \right)\,\left[ {n\left( {n+1} \right)-1} \right]+1} \right]}{42}\end{equation}

\begin{equation} \sum\limits_{k=1}^{n}{k^m}= \frac{n^{m+1}}{m+1}+\sum\limits_{j=1}^{m}{\binom {m}{j} \frac{(-1)^{j+1}}{j+1} \sum\limits_{k=1}^{n}{k^{m-j}}} \end{equation}

\begin{equation} {S_{n}}^{(m)}:=\sum\limits_{k=0}^{n}{ {k^{m}} } , \, m \in \mathbb{N_0} \end{equation}

\begin{equation} \sum\limits_{j=1}^{m}{\binom {m}{j} (-1)^j S_{n}^{(m-j)}} = (-1)^m – n^{m}   \end{equation}

\begin{equation} S_{n}^{(m)} = \frac{n^{m+1}}{m+1} + \sum\limits_{j=1}^{m}{\binom {m}{j} \frac{(-1)^{j+1}}{j+1} {S_{n}}^{(m-j)}} \end{equation}

\begin{equation} S_{n}^{(m)} = \frac{1}{m+1}{\left[n^{m+1} + \sum\limits_{j=1}^{m}{\binom{m+1}{j+1} (-1)^{j+1} S_{n}^{(m-j)}}\right]} \end{equation}

\begin{equation} \sum\limits_{j=1}^{m}{\binom {m+1}{j} {S_{n}}^{(j)}} = {{\left({n+1}\right)}^{m+1} – n – 1}  \end{equation}

\begin{equation} \sum\limits_{j=0}^{m-1}{\binom {m}{j} {S_{n}}^{(j)}} = {{\left({n+1}\right)}^{m} – \delta_{m0}}  \end{equation}

\begin{equation} {S_{n}}^{(m)} = \frac{1}{m+1} \left[{\left({{\left({n+1}\right)}^{m+1} – n – 1}\right)} – \sum\limits_{j=1}^{m-1}{\binom {m+1}{j} {S_{n}}^{(j)}}\right] \end{equation}

\begin{equation} \sum\limits_{j=0}^{m}{\binom {m-1}{j} 2^{m-j\,}S_{n}^{(j)}} = {{\left({n+2}\right)}^{m} + {\left({n+1}\right)}^{m} – \delta_{m0} – 1}  \end{equation}

\begin{equation} {S_{n}}^{(m)} = \sum\limits_{j=0}^{m}{\binom{n+1}{j+1}j!\,S(m,j)} \end{equation}

\(S(m,j)\) steht für das Stirling-Symbol zweiter Art.

\begin{equation} S(m,j) = \frac{1}{j!}\sum_{i=0}^{j}{(-1)^i\binom{j}{i}(j-i)^m} \end{equation}

\begin{equation} {S_{n}}^{(m)} = \sum\limits_{j=0}^{m} \sum_{i=0}^{j} {\binom{n+1}{j+1} {\binom{j}{i}(-1)^i\,(j-i)^m}} \end{equation}

\begin{equation} {S_{n}}^{(m)} = (n+1)\sum\limits_{j=1}^{m}{\frac{(n)_j}{(j+1)}S(m,j)}  \\ \text{wobei } (n)_j = n(n-1)(n-2)\cdots (n-j+1) \end{equation}

\begin{equation} \sum\limits_{k=1}^{n}{\binom {n}{k} {k^m}} =  \sum\limits_{j=0}^{m}{(n)_j\, 2^{n-j}S(m,j)} \\ \text{wobei } (n)_j = n(n-1)(n-2)\cdots (n-j+1) \end{equation}

\begin{equation} \sum\limits_{k=m}^{n}{(n-k)(k-m)} = \frac{(n-m-1)(n-m)(n-m+1)}{6} \end{equation}

   

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