1. Add a number to $\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+\frac{1}{210}+\frac{1}{336}$ so that the sum of the seven numbers equals $\frac{1}{4}$. What is the number to be added?
(A) $\frac{1}{56}$
(B) $\frac{1}{84}$
(C) $\frac{1}{112}$
(D) $\frac{1}{224}$
-1. C.
Add this number
$$
\begin{aligned}
= & \frac{1}{4}-\left(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+\frac{1}{210}+\frac{1}{336}\right) \\
= & \frac{1}{4}-\left(\frac{1}{1 \times 2 \times 3}+\frac{1}{2 \times 3 \times 4}+\frac{1}{3 \times 4 \times 5}+\right. \\
& \left.\frac{1}{4 \times 5 \times 6}+\frac{1}{5 \times 6 \times 7}+\frac{1}{6 \times 7 \times 8}\right) \\
= & \frac{1}{4}-\frac{1}{2}\left(\frac{1}{1 \times 2}-\frac{1}{2 \times 3}+\frac{1}{2 \times 3}-\frac{1}{3 \times 4}+\right. \\
& \frac{1}{3 \times 4}-\frac{1}{4 \times 5}+\frac{1}{4 \times 5}-\frac{1}{5 \times 6}+\frac{1}{5 \times 6}- \\
& \left.\frac{1}{6 \times 7}+\frac{1}{6 \times 7}-\frac{1}{7 \times 8}\right) \\
= & \frac{1}{4}-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{56}\right)=\frac{1}{112} .
\end{aligned}
$$