SOLUTION: Let f:R->R be a function such that f(a+b)=f(a)+f(b) and that f(2008)=3012. What is f(2009)?

From this functional equation


    f(a+b) = f(a) + f(b)


you have

     f(0) = f(0) + f(0)  ====>  f(0) = 2*f(0)  ====>  f(0) = 0,


and also  


     f(2) = 2*f(1),  f(3) = 3*f(1),  f(4) = 4*f(1),   . . . . ,  f(2008) = 2008*f(1).


Since  f(2008) = 3012 by the condition,  it implies that  2008*f(1) = 3012,  and,  hence,  f(1) =  = .



Then  f(2009) = 2009*f(1) = .


Answer.  f(2009) =  = .