What is the formula for the sum of $^{n}C_{k}$ for fixed $k$ and varying $n$? 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?
 A: Is this what you're looking for?
$$\sum_{n=k}^m {n\choose k}={m+1\choose {k+1}}$$
A: 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
$$
\sum_{j=k}^n\binom{j}{k}=\binom{n+1}{k+1}
$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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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}

$$
\color{#66f}{\large\sum_{n\ =\ n_{0}}^{n_{1}}{n \choose k}}
=\color{#66f}{\large{n_{1} + 1 \choose k + 1} - {n_{0} \choose k + 1}}
$$
