Partially hard summation

To solve this question, you must first write the general form of the fraction for the $n$ th term of the series. We have $\dfrac{1}{n(n+1)(n+2)}$ for each $n$ . We can break this up into partial fractions by doing the following:

$\dfrac{1}{n(n+1)(n+2)}$ = $\dfrac{(n+2)-n}{n(n+1)(n+2)}$ = $\displaystyle \dfrac{1}{2} (\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}).$

From this, we see that $\displaystyle \sum_{n=1}^{98} \dfrac{1}{n(n+1)(n+2)} =\dfrac{1}{2}\displaystyle[ (\dfrac{1}{1*2}-\dfrac{1}{2*3}) + (\dfrac{1}{2*3}-\dfrac{1}{3*4}) + \cdots + (\dfrac{1}{98*99}-\dfrac{1}{99*100})].$

We see that cancelation will occur, leaving us with only $\dfrac{1}{2} \displaystyle ( \dfrac{1}{1*2} - \dfrac{1}{99*100})$ = $\boxed{\dfrac{4949}{19800}}.$