2022 AMC 10B Problem 9

Adding the first few terms shows a pattern, and the pattern tells you how to split each term.

1. Small cases.
- $\frac{1}{2!} = \frac12 = 1 - \frac{1}{2!}$.
- $\frac12 + \frac{2}{3!} = \frac12 + \frac13 = \frac56 = 1 - \frac{1}{3!}$.
- $\frac56 + \frac{3}{4!} = \frac{20}{24} + \frac{3}{24} = \frac{23}{24} = 1 - \frac{1}{4!}$.

After each new term the sum is $1$ minus one over the newest factorial.
2. Why it happens. Write the top of a term as one less than the number in the factorial: $k = (k + 1) - 1$. Then
$$\frac{k}{(k+1)!} = \frac{k+1}{(k+1)!} - \frac{1}{(k+1)!} = \frac{1}{k!} - \frac{1}{(k+1)!}.$$
The last step uses $(k+1)! = (k+1) \times k!$. For example, $\frac{2}{3!} = \frac{1}{2!} - \frac{1}{3!}$, which is $\frac12 - \frac16 = \frac13$.
3. Add them all. The sum becomes
$$\left(\frac{1}{1!} - \frac{1}{2!}\right) + \left(\frac{1}{2!} - \frac{1}{3!}\right) + \cdots + \left(\frac{1}{2021!} - \frac{1}{2022!}\right).$$
Each negative piece cancels the positive piece right after it. Only the first and last pieces are left: $1 - \frac{1}{2022!}$.
4. Read off. $a = 1$ and $b = 2022$, so $a + b = 2023$.

The answer is $\boxed{\textbf{(D)}\ 2023}$.

When each term can be written as "something minus the next something", a long sum collapses because neighbours cancel. This is called telescoping. The sign to look for here is a numerator that is one less than the number in the factorial. The form $a - \frac{1}{b!}$ given in the question is a strong hint that this collapse is coming.