# Prove ${{n+1}\choose{m+1}} = \sum_{k=m}^{n}{k \choose m}$ by repeatedly using the Pascal Identity

"Prove ${{n+1}\choose{m+1}} = \sum_{k=m}^{n}{k \choose m}$ by repeatedly using the Pascal Identity ${{n+1} \choose {m+1}}={{n} \> \choose {m+1}} + {n \choose m}$"

I don't know how to start this problem. I know the version of Pascal's Identity has a 1 added to the identity. Am I just supposed to plug ${k \choose m}$ in the identity and repeat using the result?

You can use repeatedly the Pascal identity, having in mind that $n=(n-1)+1$. We have:
• Thank you! I wasn't expecting a full answer but I appreciate it. I am a bit confused as to why $n = (n - 1) + 1$ though. – Qwurticus Nov 9 '16 at 23:41
• Do you really have no idea why $n=(n-1)+1$? – Darío G Nov 9 '16 at 23:42