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I need to prove this $$\sum_{k=0}^n \binom{n+k}{k}=\binom{2n+1}{n}$$ Using induction and basically only the Pascal's identity and the symmetry property for binomial coefficients. Now, I noticed that $\sum_{k=0}^n \binom{r+k}{k}=\binom{r+n+1}{n}$ has a relatively simple proof by induction and it basically holds for any $r$. Could I apply that property for $r=n$ and call it a day?

I also tried to solve the problem by using a similar argument, the induction step however states that

$$\sum_{k=0}^{n+1} \binom{n+1+k}{k}=\sum_{k=0}^{n} \binom{n+1+k}{k}+\binom{2n+2}{n+1}$$ then $$\sum_{k=0}^{n+1} \binom{n+1+k}{k}=\sum_{k=0}^{n} \binom{n+k}{k}+\sum_{k=0}^{n} \binom{n+k}{k-1}+\binom{2n+2}{n+1}$$ Using the induction hypothesis $$\sum_{k=0}^{n+1} \binom{n+1+k}{k}= \binom{2n+1}{n}+\sum_{k=0}^{n} \binom{n+k}{k-1}+\binom{2n+2}{n+1}$$

And that's where I got stuck. Is this a dead end?

NotAMathematician
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    "Could I apply that property for $r=n$ and call it a day?" Yes! – JMoravitz Oct 23 '20 at 16:01
  • @JMoravitz I'm sorry I'm not very familiar with binomial coefficients. In what part of my attempt that could be useful? I'm aware that using the $r=n$ could solve it quickly, but I was wondering if I could find another proof. – NotAMathematician Oct 23 '20 at 16:15

2 Answers2

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This is the hockey stick Identity https://en.wikipedia.org/wiki/Hockey-stick_identity

For a combinatorial proof ...

Given $2n+1$ people, choose $n$ people from the first $n+k$ and the $(n+k+1)^{th}$ person.

Alternatively Choose $n+1$ poeple from the $2n+1$.

Donald Splutterwit
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what you have got is enough just you have to use $$\binom{{n}}{r} = \binom{n}{n-r}$$ on $\displaystyle \binom{n+k}{k-1}$ to get

$$\sum_{k=1}^{n} \binom{n+k}{n+1}=\binom{2n+1}{n+2} $$ and we are done

Aditya Dwivedi
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