How to simplify $\sum_{i=1}^{k}\binom{n + i - 1}{i}$? I tried reducing the sum to $\binom{n}{1}, \binom{n}{2}, \binom{n}{3}$ and so on but couldn't get a pattern.
Asked
Active
Viewed 306 times
3
-
Hint: Pascal's Rule. – Prasun Biswas Apr 05 '15 at 12:24
-
2Observe that $$\sum_{i=1}^2\binom{n+i-1}i=\binom n1+\binom{n+1}2=\frac{2n+(n+1)n}2=\frac{n^2+3n}2=\binom{n+2}2-1$$ – lab bhattacharjee Apr 05 '15 at 12:25
-
See $\sum_{i=0}^n\binom{i+k-1}{k-1}=\binom{n+k}{k}$ – Bart Michels Apr 08 '15 at 15:04
1 Answers
5
The problem becomes trivial on using Pascal's Rule. Using it, we have,
$$\binom{n+i-1}{i}=\binom{n+i}{i}-\binom{n+i-1}{i-1}$$
Now, substituting this into our required sum (say $S$) gives us a telescoping sum (the middle terms gets cancelled out).
$$S=\sum_{i=1}^k \binom{n+i-1}{i}=\sum_{i=1}^k \left\{\binom{n+i}{i}-\binom{n+i-1}{i-1}\right\}\\ \implies S=\binom{n+1}{1}-\binom{n}{0}+\binom{n+2}{2}-\binom{n+1}{1}+\ldots +\binom{n+k}{k}-\binom{n+k-1}{k-1}\\ \implies S=\binom{n+k}{k}-\binom{n}{0}\\ \implies \boxed{S=\dbinom{n+k}{k}-1}$$
Prasun Biswas
- 6,334