Falling factorial summands

Question

Representing the summand as a falling factorial Compute the sum

$$\sum_{k=1}^n\frac{1}{(k+1)(k+2)}$$

Answer 1

$\frac{1}{(k+1)(k+2)}=\frac{1}{k+1}-\frac{1}{k+2}$

$\sum_{k=1}^n\frac{1}{(k+1)(k+2)}=\sum_{k=1}^n(\frac{1}{k+1}-\frac{1}{k+2})$

$(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\ldots +(\frac{1}{n+1}-\frac{1}{n+2})$

$=\frac{1}{2}-\frac{1}{n+2}=\frac{n}{2(n+2)}$