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. 
 A: I believe you want to show $\binom{k+1}{r}+\binom{k+1}{r+1}=\binom{k+2}{r+1}$. It is easier to prove this combinatorically. 
Say we have $k+2$ people, and we want to pick a team $r+1$ of them. There are two options, we can either have a particular person call it $x$ in the team, in which there are now $\binom{k+1}{r}$ ways to choose the remaining people, or we cannot have $x$ in the team in which there are now $\binom{k+1}{r+1}$ ways to pick a team (we have $k+1$ to choose from because we are avoiding $x$). Hence, we have our desired identity. 
A: 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,
$$\binom{k+1}{r}+\binom{k+1}{r+1} = \binom{k+2}{r+1}.$$
A: 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 
