Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 20"
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==Problem== | ==Problem== | ||
| − | + | 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== | ||
| − | { | + | 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 16:51, 15 October 2007
Problem
The sequence 
satisfies 
.
Given that 
, then 
equals
A. 
B. 
C. 
D. 
E. 
Solution
Lets write out a couple of terms: 
. We quickly see that every third term is zero, so the answer is 
.
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 | ||
