Prove that $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} $ Prove that $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}  $
Thanks in advance, my professor asked us to this a couple weeks ago, but I was enable to get to the right answer. 
Good luck!
Here is what I got up to;
$\frac{(n+1)!}{(n-r)!(r+1)!} = \frac{(n)!}{(r)!(n-r)!} + \frac{(n)!}{(r+1)!(n-r-1)!}    $

Even though it seems a little far-fetched I will use the Binomial Theorem. The definition of number $\binom{n}{r}$ has a reason to come to exist in the development of $(x + y)^n$. So I think the natural and instructive. For all $x,y\in\mathbb{R}$ we have,
\begin{array}{rrl}
 \hspace{2cm}&( x+y)^{n+1}=     & (x+y)\cdot ( x+y)^n,
\\
\Longleftrightarrow & \sum_{k=0}^{n+1}\binom{n+1}{k}x^{(n+1)-k}y^{k}=  & (x+y)\cdot  \sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^{k},
\\
\Longleftrightarrow & \sum_{k=0}^{n+1}\binom{n+1}{k}x^{(n+1)-k}y^{k}=  &  \sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^{k+1}+ \sum_{k=0}^{n}\binom{n}{k}x^{n-k+1}y^{k}.
\end{array}
Then for all $k=1,2,\ldots n$ we have
\begin{array}{rl}
\binom{n+1}{k}x^{(n+1)-k}y^{k}=  &  \binom{n}{(k-1)}x^{n-(k-1)}y^{(k-1)+1}+ \binom{n}{k}x^{n-k+1}y^{k}.
\end{array}
For $x=1$ and $y=1$,
\begin{array}{rl}
\binom{n+1}{k}=  &  \binom{n}{(k-1)}+ \binom{n}{k},\qquad k=1,2,\ldots n.
\end{array}
Setting $k = r +1$ then
\begin{array}{rl}
\binom{n+1}{r+1}=  &  \binom{n}{r}+ \binom{n}{r+1},\qquad r=0,1,2,\ldots n-1.
\end{array}
