Question

$\sum _{k=4}^{9}\dfrac {1} {\left( k+2\right) \left( k+3\right) }$

Answer 1

$$\sum_{k=4}^{9}\frac{1}{(k+2)(k+3)}=\sum_{k=4}^{9}\left(\frac{1}{k+2}-\frac{1}{k+3}\right)$$

$$=$$

$$\left(\frac{1}{4+2}-\frac{1}{4+3}\right)+\left(\frac{1}{5+2}-\frac{1}{5+3}\right)+\left(\frac{1}{6+2}-\frac{1}{6+3}\right)+\left(\frac{1}{7+2}-\frac{1}{7+3}\right)+\left(\frac{1}{8+2}-\frac{1}{8+3}\right)+\left(\frac{1}{9+2}-\frac{1}{9+3}\right)$$

$$=$$

$$\left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{10}\right)+\left(\frac{1}{10}-\frac{1}{11}\right)+\left(\frac{1}{11}-\frac{1}{12}\right)$$

$$=$$

$$\frac{1}{6}-\frac{1}{12}$$

$$=$$

$$\frac{1}{12}$$