2006 Cyprus MO/Lyceum/Problem 20

Problem

The sequence $f:N \to R$ satisfies $f(n)=f(n-1)-f(n-2),\forall n\geq 3$. Given that $f(1)=f(2)=1$, then $f(3n)$ equals

$\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ -3\qquad\mathrm{(C)}\ 2\qquad\mathrm{(D)}\ 1\qquad\mathrm{(E)}\ 0$

Solution

Lets write out a couple of terms: \begin{align*}f(3)&=f(2)-f(1)=0& f(4)&=f(3)-f(2)=-1\\ f(5)&=f(4)-f(3)=-1& f(6)&=f(5)-f(4)=0\end{align*} We quickly see that every third term is zero, so the answer is $\mathrm{E}$.

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 19
Followed by
Problem 21
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