Answer to Question #72149 in Math for Sayem

Question

The function f(x) satisfies the equation f(x) = f(x−1)+f(x+1) for all values
of x. If f(1) =1 and f(2) = 3, what is the value of f(2013)?

Answer 1

Solution:
f(1)=1
f(2)=3
from the task: f(x)=f(x-1)+f(x+1), so f(x+1)=f(x)-f(x-1)
f(3)=f(2)-f(1)=3-1=2
f(4)=f(3)-f(2)=2-3=-1
f(5)=f(4)-f(3)=-1-2=-3
f(6)=f(5)-f(4)=-3-(-1)=-2
f(7)=f(6)-f(5)=-2-(-3)=1
f(8)=f(7)-f(6)=1-(-2)=3
f(9)=f(8)-f(7)=3-1=2
f(10)=f(9)-f(8)=2-3=-1
…
so
f(1)=f(7)
f(2)=f(8)
…
so period (T) of this function is 6.
f(2013)=335*6+3=f(3)=2
Answer:
f(2013)=2