Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 20"

Revision as of 16:55, 15 October 2007

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. $0$

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 $\mathrm{E}$ .

@@ Line 14: / Line 14: @@
 ==Solution==
-Lets write out a couple of terms: <math>f(3) = f(2) - f(1) = 0, f(4) = f(3) - f(2) = -1, f
+Lets write out a couple of terms:
-(5) = f(4) - f(3) = -1, f(6) = f(5)-f(4) = 0</math>. We quickly see that every third term is zero, so the answer is <math>\mathrm{E}</math>.
+<cmath>f(3) = f(2) - f(1) = 0</cmath>
+<cmath>f(4) = f(3) - f(2) = -1, f(5) = f(4) - f(3) = -1, f(6) = f(5)-f(4) = 0</cmath>
+We quickly see that every third term is zero, so the answer is <math>\mathrm{E}</math>.
 ==See also==