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

(Problem)
(solution)
Line 1: Line 1:
 
==Problem==
 
==Problem==
{{empty}}
+
The [[sequence]] <math>f:N \to R</math> satisfies <math>f(n)=f(n-1)-f(n-2),\forall n\geq 3</math>.
 +
Given that <math>f(1)=f(2)=1</math>, then <math>f(3n)</math> equals
 +
 
 +
A. <math>3</math>
 +
 
 +
B. <math>-3</math>
 +
 
 +
C. <math>2</math>
 +
 
 +
D. <math>1</math>
 +
 
 +
E. <math>0</math>
  
 
==Solution==
 
==Solution==
{{solution}}
+
Lets write out a couple of terms: <math>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</math>. We quickly see that every third term is zero, so the answer is <math>\mathrm{E}</math>.
  
 
==See also==
 
==See also==
 
{{CYMO box|year=2006|l=Lyceum|num-b=19|num-a=21}}
 
{{CYMO box|year=2006|l=Lyceum|num-b=19|num-a=21}}
 +
 +
[[Category:Introductory Algebra Problems]]

Revision as of 17:51, 15 October 2007

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