Write $\sum_{r=0}^{m}{ }^{n+r} C_{r}$ in the simplified form.
We know:
${ }^{n} C_{r}+{ }^{n} C_{r-1}={ }^{n+1} C_{r}$
$\sum_{r=0}^{m}{ }^{n+r} C_{r}={ }^{n} C_{0}+{ }^{n+1} C_{1}+{ }^{n+2} C_{2}+{ }^{n+3} C_{3}+\ldots+{ }^{n+m} C_{m}$
$\because{ }^{n} C_{0}={ }^{n+1} C_{0}$
$\therefore \sum_{r=0}^{m}{ }^{n+r} C_{r}={ }^{n+1} C_{0}+{ }^{n+1} C_{1}+{ }^{n+2} C_{2}+{ }^{n+3} C_{3}+\ldots+{ }^{n+m} C_{m}$
Using ${ }^{n} C_{r-1}+{ }^{n} C_{r}={ }^{n+1} C_{r}:$
$\Rightarrow \sum_{r=0}^{m}{ }^{n+r} C_{r}={ }^{n+2} C_{1}+{ }^{n+2} C_{2}+{ }^{n+3} C_{3}+\ldots+{ }^{n+m} C_{m}$
$\Rightarrow \sum_{r=0}^{m}{ }^{n+r} C_{r}={ }^{n+3} C_{2}+{ }^{n+3} C_{3}+\ldots+{ }^{n+m} C_{m}$
Proceeding in the same way:
$\sum_{r=0}^{m}{ }^{n+r} C_{r}={ }^{n+m} C_{m-1}+{ }^{n+m} C_{m}={ }^{n+m+1} C_{m}$
$\Rightarrow \sum_{r=0}^{m}{ }^{n+r} C_{r}={ }^{n+m+1} C_{m}$