2022 AMC 10B Problem 9

The general term is $\dfrac{k}{(k+1)!}$ for $k = 1, 2, \dots, 2021$. Rewrite the numerator as $(k+1) - 1$:
$$\frac{k}{(k+1)!} = \frac{k+1}{(k+1)!} - \frac{1}{(k+1)!} = \frac{1}{k!} - \frac{1}{(k+1)!}.$$

Now the sum telescopes:
$$\left(\frac{1}{1!} - \frac{1}{2!}\right) + \left(\frac{1}{2!} - \frac{1}{3!}\right) + \cdots + \left(\frac{1}{2021!} - \frac{1}{2022!}\right) = 1 - \frac{1}{2022!}.$$

So $a = 1$ and $b = 2022$, giving $a + b = 2023$.

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

A term of the form $\frac{k}{(k+1)!}$ is a difference of consecutive reciprocal factorials, exactly as $\frac{1}{k(k+1)} = \frac1k - \frac{1}{k+1}$. The signal is a numerator that is one less than the factorial's argument. The target form $a - \frac{1}{b!}$ in the problem statement is also a strong hint that a telescoping collapse is expected.

Check small cases: $\frac{1}{2!} = 1 - \frac{1}{2!}$; adding $\frac{2}{3!}$ gives $\frac{5}{6} = 1 - \frac{1}{3!}$; adding $\frac{3}{4!}$ gives $\frac{23}{24} = 1 - \frac{1}{4!}$. The pattern $1 - \frac{1}{(n+1)!}$ after $n$ terms is clear, so with $2021$ terms the sum is $1 - \frac{1}{2022!}$.

• Splitting k/(k+1)! incorrectly (for example as 1/(k+1)! − 1/k!) and getting a negative result, or misidentifying b as 2021.
• Reading the last term's factorial as $2021!$ and answering $2022$.