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

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 $\mathrm{E}$ .

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

Page

Toolbox

Search

2006 Cyprus MO/Lyceum/Problem 20

Problem

Solution

See also