In Need of a Proof

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

I recently came across this interesting equation: k=0n(m+k1k)=(n+mn)\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?

Note by Bob Krueger
6 years, 3 months ago

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5 votes

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

As an additional hint, (m10)=1=(m0){m-1 \choose 0} = 1 = {m \choose 0}.

Calvin Lin Staff - 6 years, 3 months ago

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The hockey stick identity! Thanks!

Bob Krueger - 6 years, 3 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 - 6 years, 3 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 xmx^m in k=0n(1+x)(m+k1)xmk\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 xmx^m in (1+x)(m+n)x(mn)(1+x)^{(m+n)}*x^{(m-n)} ....
which is ((m+n)n){(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 - 6 years, 3 months ago

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Really late but here goes: (Induction on n)

We establish a base case as n=0n=0:

i=00(m+i1i)=(m0)\sum_{i=0}^{0} \dbinom{m+i-1}{i} = \dbinom{m}{0}

\Rightarrow (m10)=(m0)1=1\dbinom{m-1}{0} = \dbinom{m}{0} \Rightarrow 1=1

The base case clearly holds. Now the inductive case for n=kn=k:

i=0k(m+i1i)=(m+kk)\sum_{i=0}^{k} \dbinom{m+i-1}{i} = \dbinom{m+k}{k}

\Rightarrow (m+kk+1)+i=0k(m+i1i)=(m+kk+1)+(m+kk)\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 i=0k+1(m+i1i)=(m+k)!(k+1)!(m1)!+(m+k)!k!m!\sum_{i=0}^{k+1} \dbinom{m+i-1}{i} = \frac{(m+k)!}{(k+1)!(m-1)!} + \frac{(m+k)!}{k!m!}

\Rightarrow i=0k+1(m+i1i)=(m+k)!(m+k+1)(k+1)!m!\sum_{i=0}^{k+1} \dbinom{m+i-1}{i} = \frac{(m+k)!(m+k+1)}{(k+1)!m!}

\Rightarrow i=0k+1(m+i1i)=(m+k+1k+1)\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=kn=k, then it holds for n=k+1n=k+1. The proof follows by induction.

QED

A Former Brilliant Member - 5 years, 2 months ago

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This can be easily verified by forming a recursion i hope...

Soham Chanda - 6 years, 3 months ago

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