Let f: R → R be a function defined f(x) = 2 e ^ 2x e ^ 2x + e ....

$f\left(x\right) = \dfrac{2{e}^{2x}}{{e}^{2x} + e}$

$f\left(1 - x\right) = \dfrac{2{e}^{2\left(1 - x\right)}}{{e}^{2\left(1 - x\right)} + e} = \dfrac{2e}{e + {e}^{2x}}$

Now, $f\left(x\right) + f\left(1 - x\right) = 2$

i.e. $f\left(\dfrac{1}{100}\right) + f\left(\dfrac{99}{100}\right) = 2$

$⇒ f\left(\dfrac{1}{100}\right) + f\left(\dfrac{2}{100}\right) + ...f\left(\dfrac{99}{100}\right) = 2 \times 49 + f\left(\dfrac{50}{100}\right)$

$= 98 + 1 = 99$