Solve the following

Question:

Write $\sum_{r=0}^{m}{ }^{n+r} C_{r}$ in the simplified form.

Solution:

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:

$\sum_{r=0}^{m}{ }^{n+r} C_{r}={ }^{n+m} C_{m-1}+{ }^{n+m} C_{m}={ }^{n+m+1} C_{m}$

$\Rightarrow \sum_{r=0}^{m}{ }^{n+r} C_{r}={ }^{n+m+1} C_{m}$