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2018 AMC 10B Problems/Problem 20: Difference between revisions

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Thus, <math>f\left(2018\right) = 2016 + f\left(2\right) = 2017</math>. <math>\boxed{B}</math>
Thus, <math>f\left(2018\right) = 2016 + f\left(2\right) = 2017</math>. <math>\boxed{B}</math>
==Solution 2==


==See Also==
==See Also==
{{AMC10 box|year=2018|ab=B|num-b=19|num-a=21}}
{{AMC10 box|year=2018|ab=B|num-b=19|num-a=21}}
{{MAA Notice}}
{{MAA Notice}}

Revision as of 17:24, 16 February 2018

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

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

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

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

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

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

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

See Also

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America.

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