Wacky Exponents!

It is clear from the given expression that we need to deal with the expression $\displaystyle \frac{9^x}{9^x + 3}$ . So let us consider the expression in the form of a function.

Let $\displaystyle f(x) = \frac{9^x}{9^x + 3}$ .
$\Rightarrow f(1-x) = \frac{9^{(1-x)}}{9^{(1-x)} + 3} = \frac{9}{9 + (3)(9^x)} = \frac{3}{9^x + 3}$ .

$\Rightarrow f(x) + f(1-x) = \frac{9^x}{9^x + 3} + \frac{3}{9^x + 3} = 1$ .
Also, clearly $\displaystyle f(\frac{1}{2}) = \frac{1}{2}$ .
The required sum is, $f(\frac{1}{2000}) + f(\frac{2}{2000}) + \dots + f(\frac{999}{2000}) + f(\frac{1000}{2000}) + f(\frac{1001}{2000}) + \dots + f(\frac{1999}{2000})$
$= f(\frac{1}{2000}) + f(\frac{2}{2000}) + \dots + f(\frac{999}{2000}) + f(\frac{1}{2}) + f(1-\frac{999}{2000}) + \dots + f(1-\frac{1}{2000})$ $= \sum_{r=1}^{999} (f(\frac{r}{2000}) + f(1-\frac{r}{2000})) + f(\frac{1}{2})$

$= (999)(1) +\frac{1}{2} = \frac{1999}{2} \Rightarrow \boxed{ \frac{a}{b} = \frac{1999}{2} }$ $\Rightarrow \boxed{a+b = 2001}$

first plus last term, second plus last but one, .........all equal to one. There are 999 such pair. This gives the sum 999.
One term remaining is .....{ 9^(1000/2000)} / { 9^(1000/2000) + 3 } 199= 1/2.
Thus the total is 999.5 = 1999 / 2 = a/b. ... a + b = 2001