2006 Cyprus MO/Lyceum/Problem 20

Revision as of 17:55, 15 October 2007 by Azjps (talk | contribs) (Solution: spacing)

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

A. $3$

B. $-3$

C. $2$

D. $1$

E. $0$

Solution

Lets write out a couple of terms:

\[f(3) = f(2) - f(1) = 0\] \[f(4) = f(3) - f(2) = -1, f(5) = f(4) - f(3) = -1, f(6) = f(5)-f(4) = 0\]

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
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 26 27 28 29 30