AMC 10 · 2022 · #9
Grade 7 arithmeticProblem
The sum
2!1+3!2+4!3+⋯+2022!2021 can be expressed as a−b!1, where a and b are positive integers. What is a+b?
Pick an answer.
AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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 $a + b$.
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) $2024$
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
5.NF.A.1 Step 1 - Compute the smallest partial sums.
- $S_1 = \frac{1}{2!} = \frac{1}{2}$.
- $S_2 = \frac{1}{2!} + \frac{2}{3!} = \frac{1}{2} + \frac{2}{6} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$.
- $S_3 = \frac{5}{6} + \frac{3}{4!} = \frac{5}{6} + \frac{3}{24} = \frac{20}{24} + \frac{3}{24} = \frac{23}{24}$.
💡 Three concrete numerical values give us something to look at — pattern-spotting needs data.
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: $S_n = 1 - \frac{1}{(n+1)!}$.
💡 Each partial sum lands a tiny bit short of $1$, and the 'shortage' is exactly the next factorial reciprocal.
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 $(n+1)! = (n+1) \cdot n!$).
💡 Splitting $n$ as $(n+1) - 1$ turns each term into the *difference* of two consecutive factorial reciprocals — exactly what makes a chain cancel.
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.
💡 Every inner term shows up once with a $+$ and once with a $-$ — they cancel like dominoes.
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 ✓.
💡 Two expressions in the same form match term-by-term.
4.NBT.B.4 Step 6 - Add.
- $a + b = 1 + 2022 = 2023$.
- Answer (D).
💡 Simple addition reads off the final answer.
5.NF.A.1 Compute the smallest partial sums. $S_1 = \frac{1}{2!} = \frac{1}{2}$. $S_2 = \f 5.OA.B.3 Look for a pattern. Notice $S_1 = \frac{1}{2} = 1 - \frac{1}{2}$. And $S_2 = \fr 7.EE.A.2 Find the rewrite that explains the pattern. Each numerator $n$ can be written as 7.EE.A.1 Telescope the sum. Writing each term as a difference: $S = (\frac{1}{1!} - \frac 6.EE.A.4 Match to the target form $a - \frac{1}{b!}$. We have $S = 1 - \frac{1}{2022!}$. 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.1Add 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.3Generate two numerical patterns using two given rules and identify relationships (Spotting that each $S_n$ matches $1 - \frac{1}{(n+1)!}$.)7.EE.A.2Rewrite 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.1Apply 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.4Identify when two expressions are equivalent (Matching $1 - \frac{1}{2022!}$ with $a - \frac{1}{b!}$ to read $a = 1, b = 2022$.)4.NBT.B.4Fluently 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 $a + b = 1 + 2022 = 2023$.
⭐ 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 $a + b = 1 + 2022 = 2023$.
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