Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 20"
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==Problem== | ==Problem== | ||
| − | The | + | 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 | Given that <math>f(1)=f(2)=1</math>, then <math>f(3n)</math> equals | ||
| − | + | <math>\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ -3\qquad\mathrm{(C)}\ 2\qquad\mathrm{(D)}\ 1\qquad\mathrm{(E)}\ 0</math> | |
| − | |||
| − | B | ||
| − | |||
| − | C | ||
| − | |||
| − | D | ||
| − | |||
| − | E | ||
==Solution== | ==Solution== | ||
Lets write out a couple of terms: | Lets write out a couple of terms: | ||
| − | + | <cmath> | |
| − | <cmath>f(3) = f(2) - f(1) = 0 | + | \begin{align*}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\end{align*} | ||
| + | </cmath> | ||
We quickly see that every third term is zero, so the answer is <math>\mathrm{E}</math>. | We quickly see that every third term is zero, so the answer is <math>\mathrm{E}</math>. | ||
Latest revision as of 13:33, 26 April 2008
Problem
The sequence 
satisfies 
.
Given that 
, then 
equals
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 | ||
