2018 AMC 12B Problem 18 — Solution | LIVE by Po-Shen Loh

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2018 AMC 12B Problem 18

Attempt Problem 18 of the 2018 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

18.

A function $f$ is defined recursively by $f(1)=f(2)=1$ and $f(n)=f(n-1)-f(n-2)+n$ for all integers $n\ge3.$ What is $f(2018)?$

$2016$

$2017$

$2018$

$2019$

$2020$

Answer: B

Concepts:recursion telescoping

Difficulty rating: 2150

Solution:

Repeatedly substituting the recursion into itself gives $f(n)=f(n-6)+6.$ So $f$ increases by $6$ every time $n$ increases by $6.$

Since $2018=2+6\cdot336,$ we have $f(2018)=f(2)+6\cdot336$ $=1+2016=2017.$

Thus, the correct answer is B.

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