What is the sum of number of ways of choosing n elements from (n+r) elements where r is fixed and n varies from 1 to m ? Can this be reduced to a formula ?
\sum ^m _{n=1} \binom{n + r}n
Answer
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
No comments:
Post a Comment