SAT1000 - P491

Algebra Level 3

Given that f(x)=11+xf(x)=\dfrac{1}{1+x}. {an}\{a_n\} is a sequence whose terms are all positive so that a1=1,an+2=f(an)a_1=1, a_{n+2}=f(a_n).

If a2010=a2012a_{2010}=a_{2012}, find the value of 1000(a20+a11)\lfloor 1000(a_{20}+a_{11}) \rfloor.


Have a look at my problem set: SAT 1000 problems

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