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.
Asked
Active
Viewed 69 times
2
-
By changing the summation limits to the more appropriate values $0$ and N, we get a certain sequence of integers, which can be found in the OEIS, representing $\displaystyle{2N+1\choose N+1}$. Now that we actually know what it is that we have to prove, we might as well do it by induction. – Lucian Jul 13 '14 at 17:42
1 Answers
1
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 $M$.
Start wearing purple
- 54,260