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 (Elegant and Fast)

We know \[f(3) = f(2) - f(1) + 3\] \[f(4) = f(3) - f(2) + 4\] \[f(5) = f(4) - f(3) + 5\] \[\vdots\] \[f(2018) = f(2017) - f(2016) + 2018\] Seeing all those negatives remind us of telescoping sums. This suggests we add the equations. Adding we get \[f(3) + f(4) + f(5) + \cdots + f(2018) = 3 + 4 + 5 + \cdots + 2018 - f(1) = 3 + 4 + 5 + \cdots + 2017 = \frac{(2017)(2018)}{2} - 3.\]

Notice that \[f(2018) = f(3) + f(4) + f(5) + \cdots + f(2018) - [f(3) + f(4) + f(5) + \cdots + f(2017)].\] Using the same telescoping logic we see that \[f(3) + f(4) + f(5) + \cdots + f(2017) = \frac{(2016)(2017)}{2} -3.\] So subtracting \[\frac{(2017)(2018)}{2} - 3 - \left(\frac{(2016)(2017)}{2} - 3\right) = 2017.\] This simply means the answer is $\boxed{\textbf{(B)} \text{ 2017}}.$

~coolmath_2018

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 largest odd multiple of $3$ smaller than $2018$ is $2013$, so we have \[f(2013)=2013\] \[f(2014)=2016\] \[f(2015)=2018\] \[f(2016)=2018\] \[f(2017)=2017\] \[f(2018)=\boxed{(B) 2017}.\]

Solution 3 (Bashy Pattern Finding)

Writing out the first few values, we get: $1,1,3,6,8,8,7,7,9,12,14,14,13,13,15,18,20,20,19,19...$. Examining, we see that every number $x$ where $x \equiv 1\pmod 6$ has $f(x)=x$, $f(x+1)=f(x)=x$, and $f(x-1)=f(x-2)=x+1$. The greatest number that's $1\pmod{6}$ and less $2018$ is $2017$, so we have $f(2017)=f(2018)=2017.$ $\boxed B$

Solution 4 (Algebra)

\[f(n)=f(n-1)-f(n-2)+n.\] \[f(n-1)=f(n-2)-f(n-3)+n-1.\] Adding the two equations, we have that \[f(n)=2n-1-f(n-3).\] Hence, $f(n)+f(n-3)=2n-1$. After plugging in $n-3$ to the equation above and doing some algebra, we have that $f(n)-f(n-6)=6$. Consequently, \[f(2018)-f(2012)=6.\] \[f(2012)-f(2006)=6.\] \[\ldots\] \[f(8)-f(2)=6.\] Adding these $336$ equations up, we have that $f(2018)-f(2)=6 \cdot 336$ and $f(2018)=\boxed{2017}$.

~AopsUser101


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
2018 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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 12 Problems and Solutions

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