$$\sum_{i = 1}^n \binom{i+r-1}{i}$$
I want to solve above sum combinatorially.
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I need a hint to start. – norman Oct 29 '13 at 18:23
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Using $\binom{i+r-1}i=\binom{i+r-1}{r-1}$, isn't this the same sum as here? – Martin Sleziak Oct 31 '13 at 11:52
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See $\sum_{i=0}^n\binom{i+k-1}{k-1}=\binom{n+k}{k}$ – Bart Michels Apr 08 '15 at 15:08
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Hint:
The value of ${i+r-1\choose i}+{i+r-1\choose i-1}={i+r\choose i}$ So the first couple terms look like
$${r\choose 1}+{r\choose 0}-{r\choose 0}+{r+1\choose 2}+{r+2\choose 3}+\cdots$$
$$=-{r\choose 0}+{r+1\choose 1}+{r+1\choose 2}+{r+2\choose 3}+\cdots$$
$$=-{r\choose 0}+{r+2\choose 2}+{r+2\choose 3}+\cdots$$
Can you take it from here? Note that $-{r\choose 0}=-1$ is in place to offset the addition of ${r\choose 0}$ in the first sum above.
abiessu
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after $\sum\limits_{i=1}^n {i+r \choose i}$- $\sum\limits_{i=1}^n {i+r-1 \choose i-1}$ , how we can simplify terms? and what is $ {r \choose 1}$? – norman Oct 29 '13 at 18:50
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No, use the specific terms I mention in my hint and add them directly by the definition of the values given. For example, ${r\choose 1}={r!\over 1!(r-1)!}$ as should be expected. – abiessu Oct 29 '13 at 18:54
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No, final answer is ${n+r\choose n}-1$. The last term to be added is $n+r-1\choose n$. – abiessu Oct 29 '13 at 19:31
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