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Let $$f(x)=\frac{4^x}{4^x+2}$$ and $$S=\sum_{n=1}^{2005}f\left(\frac{n}{2005}\right)$$ What is the exact value of $S$?
I tried to write $a=4^{\large\frac{1}{2005}}$, then $$S=\sum_{n=1}^{2005}\frac{a^n}{a^n+2}$$ but I still cannot simplify it. Is there any easy method?

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just see what is $f(1-x)$ – happymath Jul 29 '14 at 10:29
@happymath Why is it useful to see $f(1-x)$? I don't quite understand. – Jeel Shah Jul 29 '14 at 13:31
up vote 7 down vote accepted

Since $$f(x)+f(1-x)=\frac{4^x}{2+4^x}+\frac{4^{1-x}}{2+4^{1-x}}=\frac{4^x}{2+4^x}+\frac{2}{4^x+2}=1$$ you have: $$\begin{eqnarray*}S&=&\sum_{n=1}^{2005}f\left(\frac{n}{2005}\right)=f(1)+\sum_{i=1}^{1002}\left(f\left(\frac{n}{2005}\right)+f\left(\frac{2005-n}{2005}\right)\right)\\&=&f(1)+1002=\frac{2}{3}+1002.\end{eqnarray*}$$

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