$ n,m \in \mathbb{N} $. proof with induction to $m$:
$$\binom{n}{n} + \binom{n+1}{n} + \binom{n+2}{n} + ... + \binom{n+m}{n} = \binom{n+m+1}{n+1}$$
Try to use: for $k < n$ its $$\binom{n+1}{k+1} = \binom{n+1}{k} + \binom{n}{k}$$
Now i have to do the Start of Induction and set for $m$ = 1 because i have to do it to $m$ if im not wrong.
$$\binom{n}{n} + \binom{n+1}{n} = \binom{n+1+1}{n+1}$$
so now in short its:(i think)
$$\binom{n}{n} + \binom{n+1}{n} = \binom{n+2}{n+1}$$
and if i use the rules down i get:
$$1 + \binom{n+1}{n} = \binom{n+2}{n+1}$$
the 1 is because $$\binom{n}{n} = 1$$
but now i dont know how can i go on i hope someone can help.
down below some more Rules:
$ \binom{n}{0}=1 \quad \binom{0}{0}=1 \quad \binom{n}{1}=n \quad \binom{n}{n-1}=n \quad \binom{n}{n-1}=n$
$ \binom{n}{k}= \frac{n!}{k!(n-k)!} \quad \binom{n}{k}= \binom{n}{n-k} \quad \binom{n}{2}=\frac{n(n-1)}{2} \quad \binom{n}{3}=\frac{n(n-1)(n-2)}{6} $
Sorry for my bad english. i try to formulate all understandable.
