As I was doing a problem, I encountered this sum and I was wondering if there is a close form of this finite sum: (let $n$ be a positive integer and $r>n$)
$$\sum_{i=0}^{r} {i \choose n}$$
(where we intepret ${i\choose n}=0$ if $i< n$.)
Obviously this is equivalent to $\sum_{i=n}^{r} {i \choose n}$ but then I do not really have an idea of how to further this. I also tried to write out these terms as factorials but that seems to make this sum worse.
Could someone please help me? Thanks very much in advance!
Edit: I reckon this is actually a duplicate (link)
You have the identity
$$ \sum_{i=n}^{r}{i\choose n}={r+1\choose n+1}$$ which can be proved by induction on $r$:
For $r=n$ this is just the equality $${n\choose n}={n+1 \choose n+1}.$$ Suppose it is true for $r=k$. We get
$$ \sum_{i=n}^{k+1}{i\choose n}={k+1 \choose n} + \sum_{i=n}^{k}{i\choose n} = {k+1 \choose n} + {k+1 \choose n+1}={k+2 \choose n+1}$$ where the last equality uses a standard binomial identity.