please a doubt.......... in calculus

\hspace{-16}$Here $\bf{f(x)=\frac{x}{2009}}$ and $\bf{g(x)=\frac{f^4(x)}{(1-f(x))^4+f^4(x)}}$\\\\\\ So $\bf{g(x)=\frac{x^4}{(2009-x)^4+x^4}}$\\\\\\ So $\bf{g(k)=\frac{k^4}{(2009-k)^4+k^4}.......................(1)}$\\\\\\ $\bf{g(2009-k)=\frac{(2009-k)^4}{k^4+(2009-k)^4}................(2)}$\\\\\\ Adding $\bf{(1)+(2)}\;\;,$ We Get\\\\\\ $\bf{g(k)+g(2009-k)=1}$\\\\\\ So $\bf{\sum_{k=0}^{2009}g(k)=g(0)+g(1)+g(2)+..........+g(2009)=\frac{2009+1}{2}=1005}$\\\\\\ Using The Facts $\bf{g(k)+g(1-k)=1}$\\\\\\ So Sum of Digit in $\bf{\sum_{k=0}^{2009}=1+0+0+5=6}$

thanks for d soln @ man111 nvr thought the way