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Albert
Well-known member
- Jan 25, 2013
- 1,225
$f(5)=10$
$f(n+3)=\dfrac {f(n)-1}{f(n)+1},\,\, for \,\, all \,\,n\in N$
find f(2009)
$f(n+3)=\dfrac {f(n)-1}{f(n)+1},\,\, for \,\, all \,\,n\in N$
find f(2009)
$f(5)=10$
$f(n+3)=\dfrac {f(n)-1}{f(n)+1},\,\, for \,\, all \,\,n\in N$
find f(2009)
An equivalent and more simple statement of the problem is: given the difference equation...$f(5)=10$
$f(n+3)=\dfrac {f(n)-1}{f(n)+1},\,\, for \,\, all \,\,n\in N$
find f(2009)