Statement

For n \geq r, n, r \in \mathbb{N}:


\sum _{i = r}^n {i \choose r} = {n + 1 \choose r + 1}

Proof

Let us have a base case n = r:


{r \choose r} = {r + 1 \choose r + 1} = 1

Now suppose \sum _{i = r}^n {i \choose r} = {n + 1 \choose r + 1} for a certain n:


\sum _{i = r}^n {i \choose r} + {n + 1 \choose r} \\
= {n + 1 \choose r + 1} + {n + 1 \choose r} \\
= {n + 2 \choose r + 1}

Since n = r is true, and if a case is true for n, it is true for n + 1, this statement is true for all n \geq r.