AMC 10 · 2022 · #9

Grade 7 arithmetic

Archetype Arithmetic Expression Decomposition

factorialfraction-arithmetictelescoping-sumpattern-recognition pattern-recognitioneasier-related-problemidentify-subproblems ↑ Prerequisites: factorialfraction-arithmetic

📏 Medium solution 💡 2 insights

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 $a+b$ ?

Pick an answer.

(A)

2020

(B)

2021

(C)

2022

(D)

2023

(E)

2024

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Toolkit + CCSS Solution

Understand

Restated: The sum $\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \cdots + \frac{2021}{2022!}$ equals $a - \frac{1}{b!}$ for some positive integers $a$ and $b$. Find $a + b$.

Givens: Each term has the form $\frac{n}{(n+1)!}$ for $n = 1, 2, \ldots, 2021$; Reminder: $k! = 1 \cdot 2 \cdot 3 \cdots k$, and $(k+1)! = (k+1) \cdot k!$; Target form: sum $= a - \frac{1}{b!}$ with $a, b$ positive integers; Answer choices: (A) $2020$, (B) $2021$, (C) $2022$, (D) $2023$, (E) $2024$

Unknowns: The values of $a$ and $b$; Their sum $a + b$

Understand

Restated: The sum $\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \cdots + \frac{2021}{2022!}$ equals $a - \frac{1}{b!}$ for some positive integers $a$ and $b$. Find $a + b$.

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #5 Look for a Pattern, #7 Identify Subproblems

$2021$ terms is too many to add directly. Tool #9 (Easier Problem) — start with the first $2$ or $3$ terms, see what the partial sum looks like. Tool #5 (Pattern) — guess the closed form from the small cases and check on one more. Tool #7 (Subproblems) — break the work into 'rewrite one term cleanly' (so neighbours cancel) and 'sum the rewritten terms'. Once the term $\frac{n}{(n+1)!}$ is rewritten as a difference $\frac{1}{n!} - \frac{1}{(n+1)!}$, the whole sum telescopes — almost everything cancels.

Execute — Answer: D

#9 Solve an Easier Related Problem 5.NF.A.1 Step 1

Compute the smallest partial sums.
$S_1 = \frac{1}{2!} = \frac{1}{2}$.
$S_2 = \frac{1}{2!} + \frac{2}{3!} = \frac{1}{2} + \frac{2}{6} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$.
$S_3 = \frac{5}{6} + \frac{3}{4!} = \frac{5}{6} + \frac{3}{24} = \frac{20}{24} + \frac{3}{24} = \frac{23}{24}$.

$$S_1 = \tfrac{1}{2},\;S_2 = \tfrac{5}{6},\;S_3 = \tfrac{23}{24}$$

💡 Three concrete numerical values give us something to look at — pattern-spotting needs data.

#5 Look for a Pattern 5.OA.B.3 Step 2

Look for a pattern.
Notice $S_1 = \frac{1}{2} = 1 - \frac{1}{2}$.
And $S_2 = \frac{5}{6} = 1 - \frac{1}{6} = 1 - \frac{1}{3!}$.
And $S_3 = \frac{23}{24} = 1 - \frac{1}{24} = 1 - \frac{1}{4!}$.
Conjecture: $S_n = 1 - \frac{1}{(n+1)!}$.

$$S_1 = 1 - \tfrac{1}{2!},\;S_2 = 1 - \tfrac{1}{3!},\;S_3 = 1 - \tfrac{1}{4!}$$

💡 Each partial sum lands a tiny bit short of $1$, and the 'shortage' is exactly the next factorial reciprocal.

#7 Identify Subproblems 7.EE.A.2 Step 3

Find the rewrite that explains the pattern.
Each numerator $n$ can be written as $(n+1) - 1$.
So $\frac{n}{(n+1)!} = \frac{(n+1) - 1}{(n+1)!} = \frac{n+1}{(n+1)!} - \frac{1}{(n+1)!} = \frac{1}{n!} - \frac{1}{(n+1)!}$ (using $(n+1)! = (n+1) \cdot n!$).

$$\dfrac{n}{(n+1)!} = \dfrac{1}{n!} - \dfrac{1}{(n+1)!}$$

💡 Splitting $n$ as $(n+1) - 1$ turns each term into the *difference* of two consecutive factorial reciprocals — exactly what makes a chain cancel.

#5 Look for a Pattern 7.EE.A.1 Step 4

Telescope the sum.
Writing each term as a difference: $S = (\frac{1}{1!} - \frac{1}{2!}) + (\frac{1}{2!} - \frac{1}{3!}) + (\frac{1}{3!} - \frac{1}{4!}) + \cdots + (\frac{1}{2021!} - \frac{1}{2022!})$.
The $-\frac{1}{2!}$ from the first parenthesis cancels with the $+\frac{1}{2!}$ from the second, and so on through the chain.
Only the very first $+\frac{1}{1!}$ and the very last $-\frac{1}{2022!}$ survive.

$$S = \dfrac{1}{1!} - \dfrac{1}{2022!} = 1 - \dfrac{1}{2022!}$$

💡 Every inner term shows up once with a $+$ and once with a $-$ — they cancel like dominoes.

#7 Identify Subproblems 6.EE.A.4 Step 5

Match to the target form $a - \frac{1}{b!}$.
We have $S = 1 - \frac{1}{2022!}$.
So $a = 1$ and $b = 2022$.
Both are positive integers ✓.

$$1 - \dfrac{1}{2022!} = a - \dfrac{1}{b!} \;\Rightarrow\; a = 1,\;b = 2022$$

💡 Two expressions in the same form match term-by-term.

#9 Solve an Easier Related Problem 4.NBT.B.4 Step 6

Add.
$a + b = 1 + 2022 = 2023$.
Answer (D).

$$a + b = 1 + 2022 = 2023 \;\Rightarrow\; \textbf{(D)}$$

💡 Simple addition reads off the final answer.

[1] #9 5.NF.A.1 Compute the smallest partial sums. $S_1 = \frac{1}{2!} = \frac{1}{2}$. $S_2 = \f

[2] #5 5.OA.B.3 Look for a pattern. Notice $S_1 = \frac{1}{2} = 1 - \frac{1}{2}$. And $S_2 = \fr

[3] #7 7.EE.A.2 Find the rewrite that explains the pattern. Each numerator $n$ can be written as

[4] #5 7.EE.A.1 Telescope the sum. Writing each term as a difference: $S = (\frac{1}{1!} - \frac

[5] #7 6.EE.A.4 Match to the target form $a - \frac{1}{b!}$. We have $S = 1 - \frac{1}{2022!}$.

[6] #9 4.NBT.B.4 Add. $a + b = 1 + 2022 = 2023$. Answer (D).

Review

Reasonableness: Independent check on small $n$: for the truncated sum up to $n = 3$ the formula gives $S_3 = 1 - \frac{1}{4!} = 1 - \frac{1}{24} = \frac{23}{24}$, which matches the direct calculation in Step 1 ✓. Magnitude check: $\frac{1}{2022!}$ is astronomically small, so the sum is just shy of $1$ — that matches the partial-sum trend $\frac{1}{2}, \frac{5}{6}, \frac{23}{24}, \ldots$ heading to $1$. Choice (D) $2023$ aligns with $a = 1, b = 2022$. The neighbouring choices (C) $2022$ and (E) $2024$ would come from $b = 2021$ or $b = 2023$, both off-by-one in the telescoping endpoint.

Alternative: Tool #5 (Pattern) alone — guess $S_n = 1 - \frac{1}{(n+1)!}$ from the first three values and trust it, jumping straight to $S_{2021} = 1 - \frac{1}{2022!}$. Faster but skips the *why* (the telescoping rewrite), so it lacks the certificate. The two-tool path here gives both the answer and the algebraic proof in roughly the same time.

CCSS standards used (min grade 7)

5.NF.A.1 Add and subtract fractions with unlike denominators (Computing the small partial sums $S_1, S_2, S_3$ by adding fractions with denominators $2!, 3!, 4!$.)
5.OA.B.3 Generate two numerical patterns using two given rules and identify relationships (Spotting that each $S_n$ matches $1 - \frac{1}{(n+1)!}$.)
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem (Rewriting $\frac{n}{(n+1)!}$ as $\frac{1}{n!} - \frac{1}{(n+1)!}$ to expose the telescoping structure.)
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions (Cancelling the inner $\pm\frac{1}{k!}$ pairs across the chain of partial differences.)
6.EE.A.4 Identify when two expressions are equivalent (Matching $1 - \frac{1}{2022!}$ with $a - \frac{1}{b!}$ to read $a = 1, b = 2022$.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Final $1 + 2022 = 2023$.)

⭐ This AMC 10 problem only needs Grade 7 rewriting an expression you already know — split $\frac{n}{(n+1)!}$ into $\frac{1}{n!} - \frac{1}{(n+1)!}$, watch the chain cancel down to $1 - \frac{1}{2022!}$, so $a + b = 1 + 2022 = 2023$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 7)

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