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. 

We have $\binom{k+1}{r} = \frac{(k+1)!}{(k+1-r)!r!}$ and $\binom{k+1}{r+1} = \frac{(k+1)!}{(k+1-(r+1))!(r+1)!} = \frac{(k+1)!}{(k-r)!(r+1)!}$. Both have a common factor of $\frac{(k+1)!}{(k-r)!r!}$ so let's factor that out to get
$$\binom{k+1}{r}+\binom{k+1}{r+1} = \frac{(k+1)!}{(k-r)!r!}\left(\frac{1}{k+1-r}+\frac{1}{r+1}\right).$$
Making a common denominator, we have
$$\binom{k+1}{r}+\binom{k+1}{r+1} = \frac{(k+1)!}{(k-r)!r!}\cdot\frac{r+1+k+1-r}{(k+1-r)(r+1)}.$$
Notice we can simplify the numerator some to get $k+2$. Rewriting we have
$$\binom{k+1}{r}+\binom{k+1}{r+1} = \frac{(k+1)!}{(k-r)!r!}\cdot\frac{k+2}{((k+1)-r)(r+1)}.$$
Notice that we have $(k-r)!(k+1-r) = (k+1-r)! = ((k+2)-(r+1))!$. So we get
$$\binom{k+1}{r}+\binom{k+1}{r+1} =\frac{(k+2)(k+1)!}{((k+2)-(r+1))!(r+1)r!} = \frac{(k+2)!}{((k+2)-(r+1))!(r+1)!}.$$
Thus,
$$\binom{k+1}{r}+\binom{k+1}{r+1} = \binom{k+2}{r+1}.$$
