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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)

UnsinkableSam
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    In this case you can to this. Choose a reasonable number, say $n=5$. Compute $$\sum_{i=5}^{r} {i \choose 5}$$ for $r=5,6,7,8$. Then you will recognize the answers as binomial coefficients. – GEdgar Oct 24 '21 at 12:43

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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.

Smn
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