9.2. Given one hundred numbers: $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots, \frac{1}{100}$. We will compute 98 differences: $a_{1}=1-\frac{1}{3}, a_{2}=\frac{1}{2}-\frac{1}{4}, \ldots$, $a_{98}=\frac{1}{98}-\frac{1}{100}$. What is the sum of all these differences?
Answer: 14651/9900. Solution.
$a_{1}+a_{2}+\ldots+a_{98}=\left(1-\underset{\sim}{\frac{1}{3}}\right)+\left(\frac{1}{2}-\underset{\sim}{\frac{1}{4}}\right)+\left(\frac{1}{3}-\frac{1}{\sim}\right)+\left(\frac{1}{4}-\frac{1}{\sim}\right)+\ldots+\left(\frac{1}{\underline{97}}-\frac{1}{99}\right)+\left(\underline{\underline{\frac{1}{98}}}-\frac{1}{100}\right)$

The underlined terms cancel out with the corresponding terms of opposite sign when calculating the sum. The numbers that remain uncancelled are

$$
1+\frac{1}{2}-\frac{1}{99}-\frac{1}{100}=\frac{14651}{9900}
$$