AMC10 2022 B

AMC10 2022 B · Q9

AMC10 2022 B · Q9. It mainly tests Manipulating equations, Fractions.

The sum \[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\] can be expressed as $a-\frac{1}{b!}$, where $a$ and $b$ are positive integers. What is $a+b$?

和 \[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\] 可以表示为 $a-\frac{1}{b!}$，其中 $a$ 和 $b$ 是正整数。求 $a+b$？

(A) 2020 2020

(B) 2021 2021

(D) 2023 2023

(E) 2024 2024

Answer

Correct choice: (D)

正确答案：（D）

Solution

For all positive integers $n,$ we have \[\frac{n}{(n+1)!}=\frac{(n+1)-1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}.\] Note that the original sum is a telescoping series: \begin{align*} \frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!} &= \left(\frac{1}{1!}-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\left(\frac{1}{3!}-\frac{1}{4!}\right)+\cdots+\left(\frac{1}{2021!}-\frac{1}{2022!}\right) \\ &= \frac{1}{1!}+\left(-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+\cdots+\frac{1}{2021!}\right)-\frac{1}{2022!} \\ &= 1-\frac{1}{2022!}. \end{align*} Therefore, the answer is $1+2022=\boxed{\textbf{(D)}\ 2023}.$

对于所有正整数 $n$，有 \[\frac{n}{(n+1)!}=\frac{(n+1)-1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}.\] 原和是首项级数： \begin{align*} \frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!} &= \left(\frac{1}{1!}-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\left(\frac{1}{3!}-\frac{1}{4!}\right)+\cdots+\left(\frac{1}{2021!}-\frac{1}{2022!}\right) \\ &= \frac{1}{1!}+\left(-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+\cdots+\frac{1}{2021!}\right)-\frac{1}{2022!} \\ &= 1-\frac{1}{2022!}. \end{align*} 因此，答案是 $1+2022=\boxed{\textbf{(D)}\ 2023}$。

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