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Simple recursion (Posted on 2018-04-08) Difficulty: 2 of 5
It is given that f(x) is a function defined on R, satisfying f(1) = 1, and for any x ∈ R,
f(x+5)>f(x)+4 and f(x+1)<f(x)+2 .

If g(x)=f(x)+1-x evaluate g(2018).

No Solution Yet Submitted by Ady TZIDON    
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Some Thoughts some thoughts Comment 1 of 1
For the simplest recursion, where f(x+1)=f(x)+k, k can be anywhere from just above 4/5 to just below 2.

If k = .8, g(2018) would equal - 402.4
If k = 2, g(2018) would equal 2018

If k=1, g(2018) would be 1.

  Posted by Charlie on 2018-04-08 11:48:35
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