Does there exist a closed formula for $$\underset{n=1}{\overset{N-1}{\sum}}\dbinom{N+n}{n}?$$ I've searching on wikipedia but I haven't found this kind of sum.
2026-04-03 22:33:36.1775255616
Identity with binomials
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Yes, this sum is equal to $$ \frac{N}{N+1}{2N \choose N}-1.$$ Proof: Consider a more general identity $$\sum_{n=0}^{M-1}{N+n \choose n} =\frac{M}{N+1}{N+M \choose M}, $$ which can be proved by induction on $M$.