Figure-It-Out Function! - Part 2: Semi- A.P.
A function f : Z → Z , such that f ( n ) = f ( n − 1 ) + f ( n + 1 ) , with f ( 1 ) = 2 , and f ( 2 ) = 1 . Find f ( 2 0 1 5 )
This problem is part of the Figure-It-Out Function! series
The answer is -1.
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4 solutions
Worlds shortest solution, but still a good one.
f ( 2 0 1 5 ) = f ( 2 0 0 9 ) = f ( 2 0 0 3 ) = . . . = f ( 1 1 ) = f ( 5 )
Precise soln,
can u plz explain how f(2015)=f(5)!
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Notice that f ( n ) = f ( n − 1 ) + f ( n + 1 ) and f ( n + 1 ) = f ( n ) + f ( n + 2 ) . Add both equations and cancel terms to get f ( n + 2 ) = − f ( n − 1 ) , that is, f ( n ) = − f ( n + 3 ) . It follows that, f ( n + 3 ) = − f ( n + 6 ) , hence f ( n ) = − f ( n + 3 ) = f ( n + 6 ) . Therefore: f ( n ) = f ( n m o d 6 ) .
So f ( 2 0 1 5 ) = f ( 2 0 1 5 m o d 6 ) = f ( 5 ) = − f ( 2 ) = − 1 .
For three consecutive numbers, f(middle)= f(first) + f(last). So we find from given f(1)=2 and f(2)=1, the following.
f(1,2,3,4,5,6) ={2,1,-1,-2,-1,1} and f(n)=f(n+6).
2015=5 (mod 6)..........f(2015)=f(5)=- 1
Note:-f(2)=f(1)+f(3)..from this we get f(3)....f(3)=f(2)+f(4)..get f(4) . . .
f(1)= 2
f(2)= 1
f(3)=-1
f(4)=-2
f(5)=-1
f(6)=1
f(7)=2 .
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f(x)=f(x+6)
f(2015)=f(5)=-1
Pattern is f (0) =1 f (1) = 2 f (2) = 1 f (3) = -1 f (4) = -2 f (5) = -1 . . . . Now f (2015) = f (5) = -1
If we think about f(2015)=f(2014)-f(2013) and the we solve for f(2014) we get -f(2012) and again we solve for f(2012) just like f(2015) steps we get f(2009) so we can see a pattern that if we subtract 3 from each n we get all are equals but difference is that if we see even then - (sign) and when it comes odd we take plus sign. so the reminder of 2015 divided by 3 is equal to 2 which is even then we can say that f(2015)=-f(2) and f(2)=1 is given so final answer is -1.
Note that f ( n ) = f ( n + 6 ) and hence f ( 2 0 1 5 ) = f ( 5 ) = − 1