1. The sum
$$
\frac{1}{1 \times 2 \times 3}+\frac{1}{2 \times 3 \times 4}+\frac{1}{3 \times 4 \times 5}+\cdots+\frac{1}{100 \times 101 \times 102}
$$
can be expressed as $\frac{a}{b}$, a fraction in its simplest form. Find $a+b$,
1. Answer: 12877
Solution. Let $S$ be the required sum. By using method of difference,
$$
\begin{aligned}
S & =\frac{1}{2}\left(\frac{1}{1 \times 2}-\frac{1}{2 \times 3}+\frac{1}{2 \times 3}-\frac{1}{3 \times 4}+\cdots+\frac{1}{100 \times 101}-\frac{1}{101 \times 102}\right) \\
& =\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10302}\right) \\
& =\frac{2575}{10302}
\end{aligned}
$$

Hence $a+b=12877$.