$$a_n - a_{n-1} = - \frac{1}{n}+ \frac{1}{n + 1} + \frac{1}{n + 1} - \frac{1}{n + 2}$$Apply telescope series.

Rewrite the recursion as \begin{align*} a_n - a_{n-1} &= - \frac{1}{n}+ \frac{2}{n + 1} - \frac{1}{n + 2}\\ &= - \frac{1}{n}+ \frac{1}{n + 1} + \frac{1}{n + 1} - \frac{1}{n + 2} \end{align*}Then, \begin{align*} a_{1} &=\frac {1}{6} \\ a_{2} - a_{1} &= - \frac{1}{2}+ \frac{1}{3} + \cancel {\frac{1}{3} - \frac{1}{4}} \\ a_{3} - a_{2} &= \cancel { - \frac{1}{3}+ \frac{1}{4}} + \cancel {\frac{1}{4} - \frac{1}{5}} \\ a_{4} - a_{3} &= \cancel {- \frac{1}{4}+ \frac{1}{5} }+ \cancel {\frac{1}{5} - \frac{1}{6}} \\ ......&...... \\ a_{n-1} - a_{n-2} &= \cancel {- \frac{1}{n-1}+ \frac{1}{n}} + \cancel {\frac{1}{n} - \frac{1}{n + 1}} \\ a_{n} - a_{n-1} &= \cancel { - \frac{1}{n}+ \frac{1}{n + 1}} + \frac{1}{n + 1} - \frac{1}{n + 2} \end{align*}Add all above, and apply telescope series, \begin{align*} a_{n} &= \frac {1}{6} - \frac{1}{2}+ \frac{1}{3} + \frac{1}{n + 1} - \frac{1}{n + 2} \\ &= \frac{1}{(n+1)(n + 2)} \\ a_{100} &= \frac{1}{10302} \end{align*}