Main post link -> http://en.wikipedia.org/wiki/Multiset#Recurrence_relation

I recently came across this interesting equation: \[\displaystyle \sum_{k=0}^{n} \binom{m+k-1}{k} = \binom{n+m}{n}\] Wikipedia said that the result was related to the multiset, to which I have posted a link to in the title of the discussion. I still can't understand, though, why this equation is true. Can anyone figure it out?

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TopNewestThere's an easy direct proof, if you consider where the binomial coefficients are in Pascal's triangle.

As an additional hint, \({m-1 \choose 0} = 1 = {m \choose 0}\). – Calvin Lin Staff · 3 years, 9 months ago

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– Bob Krueger · 3 years, 9 months ago

The hockey stick identity! Thanks!Log in to reply

Really late but here goes: (Induction on n)

We establish a base case as \(n=0\):

\(\sum_{i=0}^{0} \dbinom{m+i-1}{i} = \dbinom{m}{0}\)

\(\Rightarrow\) \(\dbinom{m-1}{0} = \dbinom{m}{0} \Rightarrow 1=1\)

The base case clearly holds. Now the inductive case for \(n=k\):

\(\sum_{i=0}^{k} \dbinom{m+i-1}{i} = \dbinom{m+k}{k}\)

\(\Rightarrow\) \(\dbinom{m+k}{k+1} + \sum_{i=0}^{k} \dbinom{m+i-1}{i} =\dbinom{m+k}{k+1} +\dbinom{m+k}{k}\) (By hypothesis)

\(\Rightarrow\) \(\sum_{i=0}^{k+1} \dbinom{m+i-1}{i} = \frac{(m+k)!}{(k+1)!(m-1)!} + \frac{(m+k)!}{k!m!}\)

\(\Rightarrow\) \(\sum_{i=0}^{k+1} \dbinom{m+i-1}{i} = \frac{(m+k)!(m+k+1)}{(k+1)!m!}\)

\(\Rightarrow\) \(\sum_{i=0}^{k+1} \dbinom{m+i-1}{i} = \dbinom{m+k+1}{k+1}\)

So we may conclude that the statement holds for the base case and that if the statement holds for some \(n=k\), then it holds for \(n=k+1\). The proof follows by induction.

QED – Ethan Robinett · 2 years, 7 months ago

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Well if you need a conceptual proof here it is:

The Left hand side of the equation(LHS) can be recognized as the Coefficient of \(x^m\) in \(\sum_{k=0}^n (1+x)^{(m+k-1)}*x^{m-k}\) (Such equations can be constructed with little analysis)....

This can be treated as a geometric progression(GP) and the well know formula for a GP can be used to get the required result as coefficient of \(x^m\) in \((1+x)^{(m+n)}*x^{(m-n)}\) ....

which is \({(m+n) \choose n}\) ....

I have neglected the simplification of the GP sum you can Wikipedia for the sum of the GP formula and simplify it yourself :) – Akshay Joshi · 3 years, 9 months ago

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I guess the easiest way to prove this is to use induction, i.e., you just need to show ${m+n}\choose {n}+ {m+n}\choose {n+1}={m+n+1}\choose {n+1}$. It's not hard to show this equality using the definition of binomial expansion. As always a induction proof is not as satisfying as a conceptual explanation. – Kuai Yu · 3 years, 9 months ago

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This can be easily verified by forming a recursion i hope... – Soham Chanda · 3 years, 9 months ago

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