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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.

Remark

This question was closed for being a duplicate of How to simplify $\sum_{i=1}^{k}\binom{n + i - 1}{i}$?. However, the latter was posted in Apr 5, 2015, hence after this one.

Gonçalo
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user15765
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3 Answers3

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\sum_{k = 1}^{N}{n + k - 1 \choose k}:\ {\large ?}}$

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}

Felix Marin
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  • Could you explain the very first equality--${n + k - 1 \choose k} = \oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n + k - 1} \over z^{k + 1}} ,{\dd z \over 2\pi\ic}$, or do you have a reference? – user15765 Jul 17 '14 at 22:20
  • @user15765 The pole of $\displaystyle{\large\left(1 + z\right)^{n + k - 1} \over z^{k + 1}}$ is found by expanding the binomial in the numerator. The power $\displaystyle{\large z^{k}}$ has the coefficient $\displaystyle{\large{n + k - 1 \choose k}}$. Indeed, That's is an identity. Quite useful. – Felix Marin Jul 17 '14 at 22:27
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    @user15765 If you expand the binomial in the numerator the only contribution ( Residue Theorem ) arises from the power $\large z^{k}$. – Felix Marin Jul 17 '14 at 22:30
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    Brilliant! Thank you very much for the derivation. – user15765 Jul 17 '14 at 23:50
  • @user15765 I'm glad it was useful for you. Thanks. – Felix Marin Jul 18 '14 at 00:24
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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 $\displaystyle\sum_{k = 1}^{N}\dbinom{n+k-1}{k} = \dbinom{n+N}{n}-1$.

JimmyK4542
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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

$$ \binom{a+N+1}{N} - \binom{a+N}{0} = \binom{n+N}{N} - 1$$

Aryabhata
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