Question

Let the sum $\displaystyle \sum _{n = 1}^{9}\frac{1}{n \left(\right. n + 1 \left.\right) \left(\right. n + 2 \left.\right)}$ , written in the rational form be $\frac{p}{q}$ (where $p$ and $q$ are co-prime), then the value of $\left[\frac{q - p}{10}\right]$ is, (where [.] is the greatest integer function)

Answer 1

$\displaystyle \sum _{n = 1}^{9} \frac{1}{2} \, \frac{n + 2 - n}{n \left(n + 1\right) \, \left(n + 2\right)}$
$=\frac{1}{2} \, \left(\frac{1}{1.2} - \frac{1}{2.3}\right) + \left(\frac{1}{2.3} - \frac{1}{3.4}\right) + \ldots + \left(\frac{1}{9 \times 10} - \frac{1}{10 \times 11}\right)$
$=\frac{1}{2}\left(\frac{1}{2} - \frac{1}{110}\right)=\frac{1}{2}\left(\frac{55 - 1}{110}\right)=\frac{27}{110}$
$\Rightarrow \, \, \, q-p=110-27=83$