AMC 12 Mock Exam 5 Problem 13

Since $\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)}=\frac{(k+2)-k}{k(k+1)(k+2)}=\frac{2}{k(k+1)(k+2)}$, the sum telescopes: $$\sum_{k=1}^{20}\frac{1}{k(k+1)(k+2)}=\frac12\left(\frac{1}{1\cdot2}-\frac{1}{21\cdot22}\right)=\frac14-\frac1{924}=\frac{230}{924}=\boxed{\tfrac{115}{462}}.$$ (The distractor $\frac{209}{840}=\frac14-\frac{1}{2\cdot20\cdot21}$ comes from stopping the telescoping one term early.)