I am searching for a formula of sum of binomial coefficients $^{n}C_{k}$ where $k$ is fixed but $n$ varies in a given range? Does any such formula exist?
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Is this what you're looking for? $$\sum_{n=k}^m {n\choose k}={m+1\choose {k+1}}$$
Hypergeometricx
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As shown in this answer, we have the formula $$ \sum_{j=k}^{n-m}\binom{j}{k}\binom{n-j}{m}=\binom{n+1}{k+m+1} $$ If we set $m=0$, we get $$ \sum_{j=k}^n\binom{j}{k}=\binom{n+1}{k+1} $$
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With $\ds{n_{1} \geq n_{0}}$ and the identity $\ds{{m \choose s} \equiv \oint_{\verts{z}\ =\ 1\ }{\pars{1 + z}^{m} \over z^{s + 1}} \,{\dd z \over 2\pi\ic}\,,\quad s = 0,1,2,3,\ldots}$:
\begin{align} &\color{#66f}{\large\sum_{n\ =\ n_{0}}^{n_{1}}{n \choose k}} =\sum_{n\ =\ n_{0}}^{n_{1}}\oint_{\verts{z}\ =\ 1\ }{\pars{1 + z}^{n} \over z^{k + 1}}\,{\dd z \over 2\pi\ic} =\oint_{\verts{z}}{1 \over z^{k + 1}}\sum_{n\ =\ n_{0}}^{n_{1}}\pars{1 + z}^{n}\,{\dd z \over 2\pi\ic} \\[5mm]&=\oint_{\verts{z}\ =\ 1\ }{1 \over z^{k + 1}} \pars{1 + z}^{n_{0}}\,{\pars{1 + z}^{n_{1} - n_{0} + 1} - 1 \over \pars{1 + z} - 1}\,{\dd z \over 2\pi\ic} \\[5mm]&=\oint_{\verts{z}\ =\ 1\ }{\pars{1 + z}^{n_{1} + 1} \over z^{k + 2}} \,{\dd z \over 2\pi\ic} -\oint_{\verts{z}\ =\ 1\ }{\pars{1 + z}^{n_{0}} \over z^{k + 2}} \,{\dd z \over 2\pi\ic} \end{align}
$$ \color{#66f}{\large\sum_{n\ =\ n_{0}}^{n_{1}}{n \choose k}} =\color{#66f}{\large{n_{1} + 1 \choose k + 1} - {n_{0} \choose k + 1}} $$
Felix Marin
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Something must be wrong: when $n_0=k$ and $n_1=m$ we recover hyergeometric's answer but with an extra $-\binom k{k+1}$. – Eric Stucky Oct 19 '14 at 23:16
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1@EricStucky $\displaystyle{k \choose k + 1} = 0$. Thanks. – Felix Marin Oct 19 '14 at 23:18
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It's great how after doing combinatorics for the last six years [admittedly on and off] I still forget which number is supposed to be bigger :/ – Eric Stucky Oct 19 '14 at 23:38