Endliche Summen(4)

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(1)   \begin{equation*} \sum\limits_{k=0}^{n}{p^k} &=\frac{1 - p^{n+1}}{1-p} , \quad p\ne{1} \end{equation*}

(2)   \begin{equation*} \sum\limits_{k=0}^{n}{k\, p^k} &=p\,\frac{1-\left( {n+1} \right){p^n}+n\, p^{n+1}}{(1-p)^2} , \quad p\ne{1} \end{equation*}

(3)   \begin{equation*} \sum\limits_{i=0}^{n}{r^i}{s^{n-i}}&=\frac{r^{n+1}-{s^{n+1}}}{r-s}, \quad {r}\ne{s} \end{equation*}

(4)   \begin{align*} \sum\limits_{i+j+k=n}{r^i\,s^j\,t^k} =\frac{r^{n+2}}{\left( {r-s} \right)\left( {r-t} \right)}+\frac{s^{n+2}}{\left( {s-r} \right)\left( {s-t} \right)}+\frac{t^{n+2}}{\left( {t-s} \right)\left( {t-r} \right)} & \\r < s < t& \end{align*}

(5)   \begin{equation*} \sum\limits_{k=0}^{n} {x^{\left \lfloor {\frac{k}{p}} \right\rfloor}} &= \left({n\, - p{\left \lfloor {\frac{n}{p}} \right\rfloor}+1}\right)\,x^{\left \lfloor {\frac{n}{p}} \right\rfloor} + p\frac{x^{\left \lfloor {\frac{n}{p}}\right\rfloor} - 1}{x-1}, \quad p\ne{0},x\ne{1} \end{equation*}

(6)   \begin{equation*} \sum\limits_{k=0}^{n} {{\left \lfloor {\frac{k}{p}} \right\rfloor}x^k} &= \frac{\left({1 - x^{p\left \lfloor {\frac{n}{p}} \right\rfloor}}\right)x^p -{(1-x^p)\left \lfloor {\frac{n}{p}} \right\rfloor}x^{n+1} } {(1-x)(1-x^p)}, \quad p\ne{0},x\ne{1} \end{equation*}

(7)   \begin{equation*} \sum\limits_{k=p}^{n} {\frac{x^k}{\left \lfloor {\frac{k}{p}} \right\rfloor}} &= \frac{1}{1-x}{\left[{x^p - \frac{x^{n+1}}{\left \lfloor {\frac{n}{p}} \right\rfloor} - \sum\limits_{j=2}^{\left \lfloor {\frac{n}{p}} \right\rfloor} {\frac{x^{pj}}{j(j-1)}} }\right]}, \quad p\ge{1},x\ne{1} \end{equation*}

(8)   \begin{align*} \sum\limits_{k=1}^{n} {x^{\left \lfloor {\log_p(k)} \right\rfloor}} = (n+1){x^{\left \lfloor {\log_p(n)} \right\rfloor}} -1 + \frac{p(1-x)}{1-px} {\left({1 - {{px}^{\left \lfloor {\log_p(n)} \right\rfloor}}}\right)} &\\ p,x\in\mathbb{R},\,p > {1},\,p\cdot x <{1}& \end{align*}

(9)   \begin{align*} \sum\limits_{k=0}^{n} {x^{\left \lfloor {\sqrt{k}} \right\rfloor}} = \frac{1 + x} {(1-x)^2} + \left[n+1 - \frac{ {\left( {\left \lfloor {\sqrt{n}} \right\rfloor} x\, - {\left({{\left \lfloor {\sqrt{n}} \right\rfloor}+1}\right)}\right)}^2 + x }{(1-x)^2}\right] {x^{\left \lfloor {\sqrt{n}} \right\rfloor}} & \\ x\in\mathbb{R},\,x\ne{1} & \end{align*}

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