# 2022 AMC 10B Problems/Problem 9

## Problem

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 $a+b$? $\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024$

## Solution 1

Note that $\frac{n}{(n+1)!} = \frac{1}{n!} - \frac{1}{(n+1)!}$, and therefore this sum is a telescoping sum, which is equivalent to $1 - \frac{1}{2022!}$. Our answer is $1 + 2022 = \boxed{\textbf{(D)}\ 2023}$.

~mathboy100

## Solution 2

We add $\frac{1}{2022!}$ to the original expression $$\left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\right)+\frac{1}{2022!}=\left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2020}{2021!}\right)+\frac{1}{2021!}.$$ This sum clearly telescopes, thus we end up with $\left(\frac{1}{2!}+\frac{2}{3!}\right)+\frac{1}{3!}=\frac{2}{2!}=1$. Thus the original expression is equal to $1-\frac{1}{2022!}$, and $1+2022=\boxed{\textbf{(D)}\ 2023}$.

~not_slay (+ minor LaTeX edit ~TaeKim)

## Solution 3 (Induction)

By looking for a pattern, we see that $\tfrac{1}{2!} = 1 - \tfrac{1}{2!}$ and $\tfrac{1}{2!} + \tfrac{2}{3!} = \tfrac{5}{6} = 1 - \tfrac{1}{3!}$, so we can conclude by engineer's induction that the sum in the problem is equal to $1 - \tfrac{1}{2022!}$, for an answer of $\boxed{\textbf{(D)}\ 2023}$. This can be proven with actual induction as well; we have already established $2$ base cases, so now assume that $\tfrac{1}{2!} + \tfrac{2}{3!} + \cdots \tfrac{n-1}{n!} = 1 - \tfrac{1}{n!}$ for $n = k$. For $n = k + 1$ we get $\tfrac{1}{2!} + \tfrac{2}{3!} + \cdots \tfrac{n-1}{n!} + \tfrac{n}{(n+1)!} = 1 - \tfrac{1}{n!} + \tfrac{n}{(n+1)!} = 1 - \tfrac{n+1}{(n+1)!} + \tfrac{n}{(n+1)!} = 1 - \tfrac{1}{(n+1)!}$, completing the proof. ~eibc

## Solution 4

Let $x=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{2022!}.$

Note that \begin{align*} \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{2022!}\right)&=\frac{1}{1!}+\frac{2}{2!}+\frac{3}{3!}+\dots+\frac{2022}{2022!}\\ \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+x&=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{2021!}\\ \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+x&=x+1-\frac{1}{2022!}\\ \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)&=1-\frac{1}{2022!}. \end{align*} Therefore, the answer is $1+2022=\boxed{\textbf{(D) }2023}.$

~lopkiloinm

## Solution 5 (Combinatorics)

Suppose you are picking a permutation of $2022$ distinct elements. Suppose that the correct order of the permutation is $$x_1,x_2,x_3,\ldots,x_{2021},x_{2022}$$ We want to find the probability of picking the permutation in the wrong order.

Suppose that we have picked everything to the correct order except our last $2$ elements. That is we have $$x_1,x_2,x_3,\ldots,x_{2019},x_{2020}$$ We want pick the next element such that it does not equal to $x_{2021}$. There are $1$ ways to choose that, so we add $\frac{1}{2!}$ to the probability.

Suppose that we have picked everything to the correct order except our last $3$ elements. That is we have $$x_1,x_2,x_3,\ldots,x_{2018},x_{2019}$$ We want pick the next element such that it does not equal to $x_{2020}$. There are $2$ ways to choose that, so we add $\frac{2}{3!}$ to the probability.

This series ends up being the probability of making a permutation in the wrong order and that is of course $1-\frac{1}{2022!}$ ~lopkiloinm

## Solution 6 (Desperate)

Because the fractions get smaller, it is obvious that the answer is less than $1$, so we can safely assume that $a=1$ (this can also be guessed by intuition using similar math problems). Looking at the answer choices, $2018. Because the last term consists of $2022!$ (and the year is $2022$) we can guess that $b=2022$. Adding them yields $1+2022=\boxed{\textbf{(D) }2023}$.

~iluvme and andy_lee

## Solution 7 (Partial Fractions)

Knowing that the answer will be in the form $a-\frac{1}{b!}$, we can guess that the sum telescopes. Using partial fractions, we can hope to rewrite $\frac{n-1}{n!}$ as $\frac{A}{(n-1)!}-\frac{B}{n}$. Setting these equal and multiplying by $n!$, we get $n-1=An-B(n-1)!$. Since $An$ is the only term with $n$ with degree $1$, we can conclude that $A=1$. This means that $B=\frac{1}{(n-1)!}$. Substituting, we find that $\frac{n-1}{n!}=\frac{1}{(n-1)!}-\frac{1}{n!}$. This sum clearly telescopes and we obtain $1-\frac{1}{2022!}$. This means that our desired answer is $1+2022=\boxed{\textbf{(D) }2023}.$

~kn07

## Solution 8 (Taylor Series)

We calculate the Taylor series error to be $\frac{1}{2023!}$ and this error happens $2023$ times so it is $\frac{2023}{2023!}$ which is $\frac{1}{2022!}$

- Whiz

## Video Solution by Interstigation

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 