Question

Let $f\left(x\right)=\frac{9^x}{9^x+3}$ then find the value of the sum $f\left(\frac{1}{2008}\right)+f\left(\frac{2}{2008}\right)+f\left(\frac{3}{2008}\right)+.....+f\left(\frac{2007}{2008}\right)$

• A. $2007$
• B. $2008$
• C. $1003.5$
• D. $1004$

Answer 1

Answer: (C)

$1003.5$

$f\left(x\right)=\frac{9^x}{9^x+3}\to \left(1\right)$

$f(1-x)=\frac{9^{1-x}}{9^{1-x}+3}=\frac{3}{3+9^x}\to \left(2\right)$

$\Rightarrow f\left(x\right)+f(1-x)=\frac{9^x}{9^x+3}+\frac{3}{3+9^x}$

$\Rightarrow f\left(x\right)+f(1-x)=1\to \left(3\right)$

So,$f\left(\frac{1}{2008}\right)+f\left(\frac{2007}{2008}\right)=1$

Also, $f\left(\frac{2}{2008}\right)+f\left(\frac{2006}{2008}\right)=1$

Continuing in this manner, .

$f\left(\frac{1003}{2008}\right)+f\left(\frac{1005}{2008}\right)=1$

$f\left(\frac{1004}{2008}\right)+f\left(\frac{1004}{2008}\right)=1$

$f\left(\frac{1004}{2008}\right)=\frac{1}{2}$ .

Adding all we get

$f\left(\frac{1}{2008}\right)+f\left(\frac{2}{2008}\right)+f\left(\frac{3}{2008}\right)+...+f\left(\frac{2007}{2008}\right)$

$=1003\left(1\right)+\frac{1}{2}$

$=1003.5$