Sum of combinations with varying $n$ [duplicate]
Yes your formula is corrct and: $$\sum_{n=1}^m \dbinom{n+r}n=\sum_{n=1}^m\dbinom{n+r}{r}$$
and you can prove by induction that this $\dbinom{m+r+1}{r+1}-1$
Yes your formula is corrct and: $$\sum_{n=1}^m \dbinom{n+r}n=\sum_{n=1}^m\dbinom{n+r}{r}$$
and you can prove by induction that this $\dbinom{m+r+1}{r+1}-1$