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# In Need of a Proof

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?

Note by Bob Krueger
4 years, 7 months ago

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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, $${m-1 \choose 0} = 1 = {m \choose 0}$$.

Staff - 4 years, 7 months ago

The hockey stick identity! Thanks!

- 4 years, 7 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

- 3 years, 5 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 :)

- 4 years, 7 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.

- 4 years, 7 months ago

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

- 4 years, 7 months ago