Exercise 2. We define $a_{n}=\frac{1}{2 n-1}-\frac{1}{2 n+1}$. Thus, $a_{1}=1-\frac{1}{3}$ and $a_{2}=\frac{1}{3}-\frac{1}{5}$. Let $S=a_{1}+a_{2}+a_{3}+\cdots+a_{100}$. Calculate $201 \times S$.
Solution to exercise 2 $S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\cdots+\frac{1}{199}-\frac{1}{201}=1-\frac{1}{201}$ therefore $201 S=201-1=200$.