Show $\binom{k+1}{r}+\binom{k+1}{r+1} = \binom{k+2}{k+1} $ I have been attempting to show $$\binom{k+1}{r}+\binom{k+1}{r+1} = \binom{k+2}{r+1} $$ and my work is
$$\binom{k+1}{r}+\binom{k+1}{r+1} = \frac{(k+1)!}{r!((k+1)-r)!} + \frac{(k+1)!}{(r+1)!((k+1)-(r+1))!}$$
$$=\frac{(k+1)!(r+1)}{(r+1)!((k+1)-r)!} + \frac{(k+1)!
\color{red}{(k-r)}}{(r+1)!((k+1)-(r+1))!\color{red}{(k-r)}}$$
$$=(k+1)!\frac{k+1}{(r+1)!((k+1)-r)!}$$
which appears correct. But I am having trouble seeing the final step; where does the $(k+2)$ come from? 
Somewhere I've made a mistake but I just don't see it. 

Without assuming much, we have (just by definition): 
$\binom{n}{r}+\binom{n}{r-1}=\frac{n!}{(n-r)! r!}+\frac{n!}{(n-(r-1))! (r-1)!}$
$\frac{n!}{(n-r)! r!}+\frac{n!}{(n-(r-1))! (r-1)!}=\frac{n!}{(n-r)! r!}+\frac{n!}
{(n-r+1))! (r-1)!}$
$\frac{n!}{(n-r)! r!}+\frac{n!}
{(n-r+1))! (r-1)!}=\frac{n!}{(n-r)!(r-1)!}(\frac{1}{r}+\frac{1}{n-r+1})$
$\frac{n!}{(n-r)!(r-1)!}(\frac{1}{r}+\frac{1}{n-r+1})=\frac{n!}{(n-r)!(r-1)!}(\frac{n-r+1+r}{r.(n-r+1)})$
$\frac{n!}{(n-r)!(r-1)!}(\frac{n-r+1+r}{r.(n-r+1)})=\frac{n!}{(n-r)!(r-1)!}(\frac{n+1}{r.(n-r+1)})$
$\frac{n!}{(n-r)!(r-1)!}(\frac{n+1}{r.(n-r+1)})=\frac{(n+1)n!}{(n-r+1)(n-r)!.r(r-1)!}$
$\frac{(n+1)n!}{(n-r+1)(n-r)!.r(r-1)!}=\frac{(n+1)!}{(n-r+1)!r!}=\frac{(n+1)!}{((n+1)-r)!.r!}=\binom{n+1}{r}$
Thus, we conclude that $\binom{n}{r}+\binom{n}{r-1}=\binom{n+1}{r}$
replacing $n$ by $n+1$ we get $\binom{n+1}{r}+\binom{n+1}{r-1}=\binom{n+2}{r}$
replacing $r$ by $r+1$ we get  $\binom{n+1}{r+1}+\binom{n+1}{r}=\binom{n+2}{r+1}$
The reason i have proved for $n$ is just my comfort and nothing much :P 
