2018 AMC 10B Problem 20

Write the recursion for $n$ and for $n - 1$:
$$f(n) = f(n-1) - f(n-2) + n, \qquad f(n-1) = f(n-2) - f(n-3) + (n-1).$$
Adding them cancels $f(n-1)$ and $f(n-2)$:
$$f(n) = -f(n-3) + 2n - 1.$$
Apply this twice: $f(n-3) = -f(n-6) + 2(n-3) - 1 = -f(n-6) + 2n - 7$, so
$$f(n) = f(n-6) - 2n + 7 + 2n - 1 = f(n-6) + 6.$$

So stepping the index by $6$ raises $f$ by exactly $6$. Since $2018 = 2 + 6 \cdot 336$,
$$f(2018) = f(2) + 6 \cdot 336 = 1 + 2016 = 2017.$$

The answer is $\boxed{\textbf{(B)}\ 2017}$.

The homogeneous part $f(n) = f(n-1) - f(n-2)$ has period $6$ (its characteristic roots are sixth roots of unity), and the $+n$ forcing adds a linear drift; combining the two gives "shift by 6, add 6." Whenever a recurrence has a small period, look for a relation between $f(n)$ and $f(n - \text{period})$ and reduce the target index modulo the period.

Compute directly: $f(1), \dots, f(12) = 1, 1, 3, 6, 8, 8, 7, 7, 9, 12, 14, 14$. The differences $f(n) - n$ are $0, -1, 0, 2, 3, 2$ repeating with period $6$. Since $2018 \equiv 2 \pmod 6$, $f(2018) = 2018 + (-1) = 2017$.

• Computing a few terms, spotting a pattern in the wrong place, and concluding f(n) = n or n + 1 for all n; the offset f(n) - n cycles with period 6 and depends on n mod 6.
• Reducing $2018$ modulo $6$ incorrectly (it is $2$, not $4$) and reading off the wrong offset.