Answer
See below.
Work Step by Step
We know that ${n\choose r}=\frac{n!}{(n-r)!r!}$.
Hence here ${n\choose r}+{n\choose r+1}=\frac{n!}{(n-r)!r!}+\frac{n!}{(n-(r+1))!(r+1)!}=\frac{n!}{(n-r)!r!}+\frac{n!}{(n-r-1))!(r+1)!}=\frac{n!}{(n-r)(n-r-1)!r!}+\frac{n!}{(n-r-1))!(r+1)r!}=\frac{n!(r+1)}{(n-r)(n-r-1)!(r+1)r!}+\frac{n!(n-r)}{(n-r)(n-r-1))!(r+1)r!}=\frac{n!(r+1)+n!(n-r)}{(n-r)(n-r-1)!(r+1)r!}=\frac{n!((r+1)+(n-r))}{(n-r)!(r+1)!}=\frac{n!(n+1)}{((n+1)-(r+1))!(r+1)!}=\frac{(n+1)!}{((n+1)-(r+1))!(r+1)!}={n+1\choose r+1}$
Thus we proved what we wanted to.