Question

Let $f\left(x\right)=\frac{9^x}{9^x+3}$. Show $f\left(x\right)+f(1-x)=1$, and hence evaluate $f\left(\frac{1}{1996}\right)+f\left(\frac{2}{1996}\right)+f\left(\frac{3}{1996}\right)+...+f\left(\frac{1995}{1996}\right)$.

Answer 1

$f\left(x\right)=\frac{9^x}{9^x+3}$ (i)
and $f(1-x)=\frac{9^{1-x}}{9^{1-x}+3}$
$\Rightarrow$ $f(1-x)=\frac{\frac{9}{9^x}}{\frac{9}{9^x}+3}=\frac{9}{9+{3.9}^x}$
$f(1-x)=\frac{9}{3(3+9^x)}$ (ii)
Adding Eqs. (i) and (ii), we get
$f\left(x\right)+f(1-x)=\frac{9^x}{9^x+3}+\frac{9}{3(3+9^x)}$
$=\frac{{3.9}^x+9}{3(9^x+3)}=\frac{3(9^x+3)}{3(9^x+3)}$
$\therefore f\left(x\right)+f(1-x)=1$ (iii)
Now, putting $x=\frac{1}{1996},\frac{2}{1996},\frac{3}{1996},...,\frac{998}{1996}$ in Eq. (iii), we get
$f\left(\frac{1}{1996}\right)+f\left(\frac{1995}{1996}\right)=1$
$\Rightarrow$ $f\left(\frac{2}{1996}\right)+f\left(\frac{1994}{1996}\right)=1$
$\Rightarrow$ $f\left(\frac{3}{1996}\right)+f\left(\frac{1993}{1996}\right)=1$
..........
..........
$\Rightarrow$ $f\left(\frac{997}{1996}\right)+f\left(\frac{999}{1996}\right)=1$
$\Rightarrow$ $f\left(\frac{998}{1996}\right)+f\left(\frac{998}{1996}\right)=1$
or $f\left(\frac{998}{1996}\right)=\frac{1}{2}$
Adding all the above expression, we get
$f\left(\frac{1}{1996}\right)+f\left(\frac{2}{1996}\right)+...+f\left(\frac{1995}{1996}\right)=(1+1+1+...\text{upto}997\text{times})+\frac{1}{2}$
$=997+\frac{1}{2}$
$=997.5$