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I am asked to prove that $$ \dbinom{n-1}{0} +\dbinom{n}{1}+\dbinom{n+1}{2}+\cdots+\dbinom{n+k}{k+1}=\dbinom{n+k+1}{k+1}$$

So far what I've tried ,without looking to much at the sum I've to prove ,is that if I have $n+k+1$ blocks and I want to choose $k+1$ of them I can split $\dbinom{n+k+1}{k+1}$ in two major cases,namely $$\dbinom{n+k+1}{k+1}=\dbinom{n+k}{k+1} +\dbinom{n+k}{k}$$

This is equivalent to choose $k+1$ blocks of $n+k+1$ excluding a specific block $A$ ,$\dbinom{n+k}{k+1}$,and the other case is where I include the specific block $A$ ,$\dbinom{n+k}{k}$.

I've only matched one term,so I belive that the rest of the terms is just some identity applied to $\dbinom{n+k}{k}$ which I can't see ,or don't know, now.

Any help to put me in the right route is appreciated.

Martin Sleziak
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Mr. Y
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  • This seems to be basically the same sum as here: http://math.stackexchange.com/questions/833451/prove-sum-i-0n-binomik-1k-1-binomnkk – Martin Sleziak Jan 18 '16 at 11:19
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    When you apply your identiy to RHS and simplify, you obtain the same problem but one step smaller. So you could continue to do the reduction. – Mok-Kong Shen Jan 19 '16 at 11:12

3 Answers3

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LHS can be written as:

$$\binom{n-1}0+\left[\binom{n}1-\binom{n-1}0\right]+\cdots+\left[\binom{n+k+1}{k+1}-\binom{n+k}{k}\right]$$

"Telescoping" you find the RHS.

drhab
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  • Can you elaborate a little bit ?I don't understand the pattern in the subtracting and if this is applied to my equation or the one I have to prove. – Mr. Y Jan 18 '16 at 10:30
  • Do you agree that the the first two terms of the expression in my answer sum up to $\binom{n}1$? That is the first step. It is like $a_0+(a_1-a_0)+(a_2-a_1)+\cdots=a_1+(a_2-a_1)+\cdots=a_2+\cdots$ etc. If the last term is $a_{k+1}-a_k$ then you will end up with $a_{k+1}$ as outcome. – drhab Jan 18 '16 at 10:36
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Suppose you have $n$ indistinguible balls and $k$ diferent boxes. The number of ways for distribute the $n$ balls in the $k$ boxes is $\binom{n+k-1}{k-1}$.

On the other hand, we can first put $r$ balls in the first box, $r\in\{0,1,...,n\}$, and distribute the other $n-r$ in the $k-1$ boxes: $\binom{n-r+k-1-1}{k-2}=\binom{n-r+k-2}{k-2}$.

Since the two ways of distributed the balls are equivalents, then $\binom{n+k-1}{k-1}=\sum_{r=0}^n\binom{n-r+k-2}{k-2}$

Using it, it is easy to proof your assertion.

PD: The identity you try to prove is known as "Golf Stick Identity".

sinbadh
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I think the left-hand side should have $k+2$ terms, not four: $$\dbinom{n-1}{0} +\dbinom{n}{1}+\dbinom{n+1}{2}+\cdots+\dbinom{n+k}{k+1}=\dbinom{n+k+1}{k+1}$$ The rest of the terms after you subtract $n+k\choose k+1$ have the same form as the original equation, but with $k$ replaced by $k-1$. So use induction on $k$.

Empy2
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