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I have been quite stuck on this problem and with the help of others, I may have solved it. This is what I have to prove. $$\binom{N}{n+1}=\sum^{N-1}_{k=n} \binom{k}{n}.$$

So far I have this
$${N \choose {n+1}}={{N-1}\choose{n+1}}+{{N-1}\choose n}.$$ by Pascals triangle. If I repeat this idea, I believe I end up with:
$$={{N-x}\choose{n+1}}+{{N-x}\choose{n}}...+{{{N-1}\choose n}}$$For some $x$ such that $x=n+1$. But lets go one further. $$={{N-x-1}\choose{n+1}}+{{N-x-1}\choose{n}}...+{{{N-1}\choose n}}.$$
But this reduces to $${{n}\choose{n+1}}+{{n}\choose{n}}...+{{{N-1}\choose n}}.$$ However $${{n}\choose{n+1}}=0$$. Therefore we are just left with: $\sum^{N-1}_{k=n} \binom{k}{n}.$ as desired.

Jack Armstrong
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1 Answers1

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Prove by induction that, if $\,r\le N-n-2$: $$\binom N{n+1}=\sum_{i=0}^{r}\binom {N-1-i}n +\binom{N-1-r}{n+1}$$ Indeed, for $r=0$, it is Pascal's formula. If $r<N-n-2$, use again the formula in the last term: \begin{align*} \binom N{n+1}&=\sum_{i=0}^{r}\binom {N-1-i}n +\binom{N-1-r-1}{n}+\binom{N-1-r-1}{n+1}\\&=\sum_{i=0}^{r+1}\binom {N-1-i}n + \binom{N-1-r-1}{n+1} \end{align*}

Now setting $r$ to the maximum value $N-n-2$, you get: \begin{align*} \binom N{n+1}&=\sum_{i=0}^{N-n-2}\binom {N-1-i}n +\binom{n+1}{n+1}\\ &=\sum_{i=0}^{N-n-2}\binom {N-1-i}n +\binom nn=\sum_{i=0}^{N-n-1}\binom {N-i}n \end{align*}

Then set $k=N-1-i\,$ to get the formula.

Bernard
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