Funny functions

Note first that

$\dfrac{4^{1-x}}{4^{1-x} + 2} + \dfrac{4^{x}}{4^{x} + 2} = \dfrac{4^{1-x}(4^{x} + 2) + 4^{x}(4^{1-x} + 2)}{(4^{1-x} + 2)(4^{x} + 2)} = \dfrac{4 + 4 + 2(4^{1 - x} + 4^{x})}{4 + 4 + 2(4^{1-x} + 4^{x})} = 1.$

Thus $f(x) + f(1-x) = 1$ for all real $x$ . So in the given sum, we can pair terms of the form $f(\frac{k}{2009}) + f(\frac{2009 - k}{2009}) = 1$ from $k = 1$ to $k = 1004$ to get a final sum of $\boxed{1004}.$

What I did was first find the sum of all the numbers between 2008 and 1. For that I just used the formula $\frac { n(n+1) }{ 2 }$ I realized is would work because: 2008+1=2009 2007+ 2=2009 2006+3=2009 And it continues in this pattern. I knew that there would be half the amount of pairs than the number (2008). $\frac { n }{ 2 }$ . Then I multiplied by the value of each pair, 2009 or n+1 to get $\frac { n(n+1) }{ 2 }$ Which simplifies to 1004 after you subsitute 2008 for n