Difference between revisions of "2018 AMC 10B Problems/Problem 20"

(Solution)
m (Solution 2)
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==Solution 2==
 
==Solution 2==
Let us write down the first couple terms of the sequence:
+
Choose <math>/sqare[B]</math> for B%$^^&
<math>f(1)=1</math>
 
<math>f(2)=1</math>
 
$f(3)
 
  
 
==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 18:13, 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, $f\left(2018\right) = 2016 + f\left(2\right) = 2017$. $\boxed{B}$

Solution 2

Choose $/sqare[B]$ for B%$^^&

See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
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

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