In Need of a Proof

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?

Easy Math Editor

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Comments

There'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 - 7 years, 11 months ago

The hockey stick identity! Thanks!

Bob Krueger - 7 years, 11 months ago

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 - 7 years, 11 months ago

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 - 7 years, 11 months ago

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

A Former Brilliant Member - 6 years, 10 months ago

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

Soham Chanda - 7 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$