AMC 10 · 2022 · #9

Grade 7 arithmetic
Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps
factorialfraction-arithmetictelescoping-sumpattern-recognition pattern-recognitioneasier-related-problemidentify-subproblems ↑ Prerequisites: factorialfraction-arithmetic
📏 Medium solution 💡 2 insights
Problem
The sum 12!\frac{1}{2!} + 23!\frac{2}{3!} + 34!\frac{3}{4!} + … + 20212022!\frac{2021}{2022!} equals a - 1b!\frac{1}{b!} for some positive integers a and b. Find a + b.

Pick an answer.

(A)
2020
(B)
2021
(C)
2022
(D)
2023
(E)
2024

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Solve an Easier Related Problem

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 n(n+1)!\frac{n}{(n+1)!} is rewritten as a difference 1n!\frac{1}{n!} - 1(n+1)!\frac{1}{(n+1)!}, the whole sum telescopes — almost everything cancels.

1STEP 1

Compute the first three partial sums: S₁ = 12\frac{1}{2}, S₂ = 56\frac{5}{6}, S₃ = 2324\frac{23}{24}.

S₁ = 12\frac{1}{2}, S₂ = 56\frac{5}{6}, S₃ = 2324\frac{23}{24}
2STEP 2

Spot the pattern: each Sₙ falls short of 1 by one factorial reciprocal, so conjecture Sₙ = 1 - 1(n+1)!\frac{1}{(n+1)!}.

S₁ = 1 - 12!\frac{1}{2!}, S₂ = 1 - 13!\frac{1}{3!}, S₃ = 1 - 14!\frac{1}{4!}
3STEP 3

Split n = (n+1) - 1 to rewrite each term: n(n+1)!\frac{n}{(n+1)!} = 1n!\frac{1}{n!} - 1(n+1)!\frac{1}{(n+1)!}.

n(n+1)!\frac{n}{(n+1)!} = 1n!\frac{1}{n!} - 1(n+1)!\frac{1}{(n+1)!}
4STEP 4

The differences telescope — inner terms cancel in pairs, leaving S = 1 - 12022!\frac{1}{2022!}.

S = 11!\frac{1}{1!} - 12022!\frac{1}{2022!} = 1 - 12022!\frac{1}{2022!}
5STEP 5

Match 1 - 12022!\frac{1}{2022!} to a - 1b!\frac{1}{b!}: a = 1, b = 2022, both positive integers.

1 - 12022!\frac{1}{2022!} = a - 1b!\frac{1}{b!} → a = 1, b = 2022
6STEP 6

Add: a + b = 1 + 2022 = 2023, choice (D).

a + b = 1 + 2022 = 2023 → (D)
Answer
2023
Independent check on small n: for the truncated sum up to n = 3 the formula gives S₃ = 1 - 14!\frac{1}{4!} = 1 - 124\frac{1}{24} = 2324\frac{23}{24}, which matches the direct calculation in Step 1 ✓. Magnitude check: 12022!\frac{1}{2022!} is astronomically small, so the sum is just shy of 1 — that matches the partial-sum trend 12\frac{1}{2}, 56\frac{5}{6}, 2324\frac{23}{24}, … 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.
💡Key takeaway

This AMC 10 problem only needs Grade 7 rewriting an expression you already know — split n(n+1)!\frac{n}{(n+1)!} into 1n!\frac{1}{n!} - 1(n+1)!\frac{1}{(n+1)!}, watch the chain cancel down to 1 - 12022!\frac{1}{2022!}, so a + b = 1 + 2022 = 2023.

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