AMC 10 · 2024 · #21
Grade 9 algebraPick an answer.
The recurrence is hostile in its printed form: the coefficients change with n, so none of the standard "constant coefficient" tricks apply. Tool #15 (Organize Information in More Ways) fixes the surface by clearing both denominators, which turns the proportion into na_n - (n-1)a_n-1 = 2n - 1. That rearranged line is the hint: the same expression (index) × (term) shows up at n and at n-1. Tool #4 (Introduce a Variable) acts on the hint by naming that expression b_n = na_n, and the recurrence collapses to b_n - b_n-1 = 2n-1 — consecutive differences, which telescope in one line. Tool #5 (Look for a Pattern) runs alongside as a cheap safety net: computing a₂, a₃, a₄ by hand takes under a minute and predicts the closed form before any algebra, so the algebra has something to agree with. Once a_n = n + 1/n is in hand the sum splits into three pieces, two of which are exact; the third, Σ 1/n², has no elementary closed form, so tool #9 (Solve an Easier Related Problem) swaps it for a slightly larger series that telescopes and traps it strictly between 1 and 2. Tool #3 (Eliminate Possibilities) closes the problem: the choices are consecutive integers, and a bracket of width 1 knocks out four of them.
Generate the first few terms
The pattern looks like n plus one over n.
A sequence is a function of n, so writing each term over the denominator n lets the rule for n show itself.
9.F-IF.A.3Look For A PatternClear both denominators
It becomes a step of two n minus one.
Clearing the fractions is not cleanup for its own sake — it is what lets the two sides show the same pattern at n and n-1.
7.EE.A.1Organize Information In More WaysName the product
The new sequence adds odd numbers.
When a recurrence links na_n to (n-1)a_n-1, the thing worth naming is the product, not the term.
9.F-BF.A.1Introduce A VariableRead off the closed form
The guess holds: n plus one over n.
Splitting (n²+1)/n into n + 1/n separates the big part of the term from the tiny part, and only the tiny part will be hard to sum.
Splitting the term into a big part and a tiny part separates the sum that can be computed from the one that cannot.
▸ Why?
A sum can be broken apart term by term, so each piece can be handled on its own.
▸ Why?
The tiny part never has to be computed exactly, only trapped between two numbers you can compute.
Square and split the sum
The sum splits into three pieces.
Squaring n + 1/n makes the cross term collapse to a constant, which is why the sum breaks apart so cleanly.
6.EE.A.3Organize Information In More WaysEvaluate the two exact pieces
Those two give 338,550.
Two of the three pieces are exact integers, so the only thing left to worry about is how big the third piece is.
6.EE.A.2Organize Information In More WaysTrap the leftover tail
The tail is trapped between 1 and 2.
A sum you cannot compute can still be pinned down by a sum you can, as long as the easier one is always bigger.
9.A-SSE.A.2Solve An Easier Related ProblemTake the floor
So the floor is 338,551.
Once a number is trapped strictly between two consecutive integers, its floor is settled without ever computing it exactly.
6.NS.C.7Eliminate PossibilitiesMultiply the recurrence out and it says na_n - (n-1)a_n-1 = 2n-1, so the product na_n telescopes to n²+1 and a_n = n + 1/n; squaring gives n² + 2 + 1/n², and since the leftover Σ 1/n² is stuck strictly between 1 and 2, the floor is 338,551.
- Generate the first few terms
- Clear both denominators
- Name the product na_n
- Read off the closed form
- Square and split the sum
- Evaluate the two exact pieces
- Trap the leftover tail
- Take the floor