Let f : R $\to$ R be a function defined by $f(x) = {{2{e^{2x}}} \over {{e^{2x}} + e}}$. Then $$f\left( {{1 \over {100}}} \right) + f\left( {{2 \over {100}}} \right) + f\left( {{3 \over {100}}} \right) + \,\,\,.....\,\,\, + \,\,\,f\left( {{{99} \over {100}}} \right)$$ is equal to ______________.

<p>Given,</p> <p>$f(x) = {{2{e^{2x}}} \over {{e^{2x}} + e}}$</p> <p>$\therefore$ $f(1 - x) = {{2{e^{2(1 - x)}}} \over {{e^{2(1 - x)}} + e}}$</p> <p>$= {{2\,.\,{{{e^2}} \over {{e^{2x}}}}} \over {{{{e^2}} \over {{e^{2x}}}} + e}}$</p> <p>$= {{2{e^2}} \over {{e^2} + {e^{2x}}\,.\,e}}$</p> <p>$= {{2{e^2}} \over {e(e + {e^{2x}})}}$</p> <p>$= {{2e} \over {e + {e^{2x}}}}$</p> <p>$\therefore$ $$f(x) + f(1 - x) = {{2{e^{2x}}} \over {{e^{2x}} + e}} + {{2e} \over {{e^{2x}} + e}}$$</p> <p>$= {{2({e^{2x}} + e)} \over {{e^{2x}} + e}}$</p> <p>$= 2$ ...... (1)</p> <p>Now,</p> <p>$f\left( {{1 \over {100}}} \right) + f\left( {{{99} \over {100}}} \right)$</p> <p>$= f\left( {{1 \over {100}}} \right) + f\left( {1 - {1 \over {100}}} \right)$</p> <p>$= 2$ [as $f(x) + f(1 - x) = 2$]</p> <p>Similarly,</p> <p>$f\left( {{2 \over {100}}} \right) + f\left( {1 - {2 \over {100}}} \right) = 2$</p> <p>$\vdots$</p> <p>$$f\left( {{{49} \over {100}}} \right) + f\left( {1 - {{49} \over {100}}} \right) = 2$$</p> <p>$\therefore$ Total sum $= 49 \times 2$</p> <p>Remaining term $= f\left( {{{50} \over {100}}} \right) = f\left( {{1 \over 2}} \right)$</p> <p>Put $x = {1 \over 2}$ in equation (1), we get</p> <p>$f\left( {{1 \over 2}} \right) + f\left( {1 - {1 \over 2}} \right) = 2$</p> <p>$\Rightarrow 2f\left( {{1 \over 2}} \right) = 2$</p> <p>$\Rightarrow f\left( {{1 \over 2}} \right) = 1$</p> <p>$\therefore$ Sum $= 49 \times 2 + 1 = 99$</p>