Sum of $\frac{25^x}{25^x+5}$

Question

Find $$\sum_{n=1}^{1999}\frac{25^{\frac{n}{2000}}}{25^{\frac{n}{2000}}+5}$$

My work: $$\frac{25^x}{25^x+5}=\frac{5^{2x}}{5^{2x}+5 }$$

I don't see a pattern here, and I don't think the point of this problem is to evaluate radicals. Please help!

Answer 1

Let $f(x)=\frac{25^{x}}{25^{x}+5}.$ Note that$$f(x)+f(1-x)=\dfrac{25^{x}}{25^{x}+5}+\dfrac{25^{1-x}}{25^{1-x}+5}=1.$$ So we have$$\begin{aligned} \sum_{n=1}^{1999}f(\dfrac{n}{2000})&=[f(\frac{1}{2000})+f(\frac{1999}{2000})]+\cdots\\ &+[f(\frac{999}{2000})+f(\frac{1001}{2000})]+f(\frac{1}{2})\\ &=999+\dfrac{1}{2}=999.5.\\ \end{aligned}$$