Prove $\sum\limits_{i=0}^{n}\binom{n+i}{i}=\binom{2n+1}{n+1}$ I'm trying to prove this algebraically: $$\sum\limits_{i=0}^{n}\dbinom{n+i}{i}=\dbinom{2n+1}{n+1}$$
Unfortunately I've been stuck for quite a while. Here's what I've tried so far:


*

*Turning $\dbinom{n+i}{i}$ to $\dbinom{n+i}{n}$

*Turning $\dbinom{2n+1}{n+1}$ to $\dbinom{2n+1}{n}$

*Converting binomial coefficients to factorial form and seeing if anything can be cancelled.

*Writing the sum out by hand to see if there's anything that could be cancelled.


I end up being stuck in each of these ways, though. Any ideas? Is there an identity that can help me?
 A: Hint:
Use
$$
\binom{n+1+i}{i} - \binom{n+i}{i-1} = \binom{n+i}{i}
$$
A: Oh, hey! I just figured it out. Funny how simply posting the question allows you re-evaluate the problem differently...
So on Wikipedia apparently this is a thing (the recursive formula for computing the value of binomial coefficients): $$\dbinom{n}{k} = \dbinom{n-1}{k-1} + \dbinom{n-1}{k}$$
On the RHS of the equation we have (let's call this equation 1):
$$\dbinom{2n+1}{n+1} = \dbinom{2n}{n} + \dbinom{2n-1}{n} + \dbinom{2n-2}{n} + \cdots + \dbinom{n+1}{n} + \dbinom{2n-(n-1)}{n+1} $$
The last term $\dbinom{2n-(n-1)}{n+1}$ simplifies to $\dbinom{n+1}{n+1}$ or just 1.
Meanwhile on the LHS we have $\sum\limits_{i=0}^{n}\dbinom{n+i}{i}$ which is also $\sum\limits_{i=0}^{n}\dbinom{n+i}{n}$ because $\dbinom{n}{k} = \dbinom{n}{n-k}$.
Written out that is (let's call this equation 2):
$$\sum\limits_{i=0}^{n}\dbinom{n+i}{n} = \dbinom{n}{n} + \dbinom{n+1}{n} + \dbinom{n+2}{n} + \cdots + \dbinom{2n}{n} $$
The first term $\dbinom{n}{n}$ simplifies to 1.
Hey, look at that. 
In each written out sum there's a term in equation 1 and an equivalent in equation 2. And in each sum there's $n$ terms, so equation 1 definitely equals equation 2. I mean, the order of terms is flipped, but whatever. Yay!
