using partial fractions ...

$\displaystyle \sum_{n=1}^{100} \frac{1}{n(n+1)(n+2)} = \frac{1}{2} \sum_{n=1}^{100} \frac{1}{n} - \sum_{n=1}^{100} \frac{1}{n+1} + \frac{1}{2} \sum_{n=1}^{100} \frac{1}{n+2}$

$\displaystyle \frac{1}{2}\bigg[1+\frac{1}{2}+\frac{1}{3}+ \, ... \, + \frac{1}{99} + \frac{1}{100}\bigg] - \bigg[\frac{1}{2}+\frac{1}{3}+ \frac{1}{4} \, ... \, + \frac{1}{100} + \frac{1}{101}\bigg] + \frac{1}{2}\bigg[\frac{1}{3}+\frac{1}{4}+ \, ... \, + \frac{1}{101} + \frac{1}{102}\bigg]$

multiply each summation by 2 ...

$\displaystyle \bigg[1+\frac{1}{2}+\frac{1}{3}+ \, ... \, + \frac{1}{99} + \frac{1}{100}\bigg] - \bigg[\frac{2}{2}+\frac{2}{3}+ \frac{2}{4} \, ... \, + \frac{2}{100} + \frac{2}{101}\bigg] + \bigg[\frac{1}{3}+\frac{1}{4}+ \, ... \, + \frac{1}{101} + \frac{1}{102}\bigg]$

after a whole lot of zero cancellations, the sum of the above is $\bigg[1-\dfrac{1}{2}-\dfrac{1}{101}+\dfrac{1}{102}\bigg]$ ... all that's left is to divide this result by 2