Closed expression for a $\sum\limits_{i=1}^{m} \binom{n+i}{n}$

Asked Apr 02 '23 at 00:56

Active Apr 02 '23 at 01:40

Viewed 52 times

Find a closed expression of a following sum ($n,m \in N$):

$$\sum\limits_{i=1}^{m} \binom{n+i}{n}$$

Tried to use Pascal Identity but sum continues to grow with more complex subsums:

$$\sum\limits_{i=1}^{m}\binom{n+i}{n}=m\binom{n+1}{n}+\sum\limits_{i=1}^{m-1}(m-i)\binom{n+i}{n-1}$$

edited Apr 02 '23 at 01:40

Sil

asked Apr 02 '23 at 00:56

ramazan793

Nice question! Can you define n? Is n defined to be any natural number? – ktm5124 Apr 02 '23 at 01:02
@ktm5124 Thanks! Yes, n is a natural number – ramazan793 Apr 02 '23 at 01:10
6

You can use the hockey-stick identity to evaluate this sum. – N. F. Taussig Apr 02 '23 at 01:15
1

Mathematica calculation shows that the answer is $\frac{(m+1) \binom{m+n+1}{n}}{n+1}-1$ – 138 Aspen Apr 02 '23 at 01:57

0 Answers0