2018 AMC 10B Problems/Problem 20

Problem

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 $n \geq 3$ . What is $f(2018)$ ?

$\textbf{(A)} \text{ 2016} \qquad \textbf{(B)} \text{ 2017} \qquad \textbf{(C)} \text{ 2018} \qquad \textbf{(D)} \text{ 2019} \qquad \textbf{(E)} \text{ 2020}$

Solution 1

$f\left(n\right) = f\left(n - 1\right) - f\left(n - 2\right) + n$

$= \left(f\left(n - 2\right) - f\left(n - 3\right) + n - 1\right) - f\left(n - 2\right) + n$

$= 2n - 1 - f\left(n - 3\right)$

$= 2n - 1 - \left(2\left(n - 3\right) - 1 - f\left(n - 6\right)\right)$

$= f\left(n - 6\right) + 6$

Thus, $f\left(2018\right) = 2016 + f\left(2\right) = 2017$ . $\boxed{B}$

Solution 2 (A Bit Bashy)

Start out by listing some terms of the sequence. $f(1)=1$ $f(2)=1$

$f(3)=3$ $f(4)=6$ $f(5)=8$ $f(6)=8$ $f(7)=7$ $f(8)=7$

$f(9)=9$ $f(10)=12$ $f(11)=14$ $f(12)=14$ $f(13)=13$ $f(14)=13$

$f(15)=15$ $.....$ Notice that $f(n)=n$ whenever $n$ is an odd multiple of $3$ , and the pattern of numbers that follow will always be +3, +2, +0, -1, +0. The closest odd multiple of $3$ to $2018$ is $2013$ , so we have $f(2013)=2013$ $f(2014)=2016$ $f(2015)=2018$ $f(2016)=2018$ $f(2017)=2017$ $f(2018)=\boxed{2017}.$

Written By: pi3141592

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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2018 AMC 10B Problems/Problem 20

Contents

Problem

Solution 1

Solution 2 (A Bit Bashy)

See Also