Sum of $k$-combination with repetitions I can see that there are $\binom{n+k-1}{k}$ cases of choosing k items of n types with repetition from http://en.wikipedia.org/wiki/Multiset_coefficient#Counting_multisets.
I wonder whether there is any formula about its sum for varying $k$.
In particular, I am interested in $\sum_{k=1}^{N} \binom{n+k-1}{k}$.
Does anyone have an idea, or a link about this or anything similar to this one?
Update: I changed $nk$ to $N$. Actually $N = np$ for some positive integer $p$, but I think this is not necessary.
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$\ds{\sum_{k = 1}^{N}{n + k - 1 \choose k}:\ {\large ?}}$

The binomial $\ds{{n + k - 1 \choose k}}$ is non-zero whenever
  $\ds{0\ \leq\ k\ \leq\ n + k - 1\ \imp\ k \geq 0\,,\ n \geq 1}$. Hereafter, we'll assume those conditions are satisfied.

\begin{align}&\color{#66f}{\large\sum_{k = 1}^{N}{n + k - 1 \choose k}}
=\sum_{k = 1}^{N}\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n + k - 1} \over z^{k + 1}}
\,{\dd z \over 2\pi\ic}
\\[2mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n - 1} \over z}\sum_{k = 1}^{N}
\pars{1 + z \over z}^{k}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n - 1} \over z}
{1 + z \over z}{\bracks{\pars{1 + z}/z}^{N} - 1 \over \pars{1 + z}/z - 1}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n} \over z}
\bracks{{\pars{1 + z}^{N} \over z^{N}} - 1}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n + N} \over z^{N + 1}}
\,{\dd z \over 2\pi\ic}
-\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n} \over z}\,{\dd z \over 2\pi\ic}
=\color{#66f}{\large{N + n \choose N} - 1}
\end{align}
A: The Hockeystick Identity states that for positive integers $n>r$, we have $\displaystyle\sum^n_{i=r}{i\choose r}={n+1\choose r+1}$.
See that link for several proofs.
Thus, $\displaystyle\sum_{k = 0}^{N}\dbinom{n+k-1}{k} = \displaystyle\sum_{k = 0}^{N}\dbinom{n+k-1}{n-1} \overset{i = n+k-1}{=}\displaystyle\sum_{i = n-1}^{n+N-1}\dbinom{i}{n-1} = \dbinom{n+N}{n}$. 
Subtract $\dbinom{n+k-1}{0} = 1$ from both sides to get $\displaystyle\sum_{k = 1}^{N}\dbinom{n+k-1}{k} = \dbinom{n+N}{n}-1$. 
A: This can be written as a telescopic sum (with $a = n-1$)
$$\sum_{k=1}^{N} \binom{a+k}{k} = \sum_{k=1}^{N} \left(\binom{a+k+1}{k} - \binom{a+k}{k-1}\right)$$
which gives us the answer to be
$$ \binom{a+N+1}{N} - \binom{a+N}{0} = \binom{n+N}{N} - 1$$
