Question

Let $f\left(x\right)=\frac{3^x}{\sqrt{3}+3^x}$ and ${\sum }_{r=0}^{2n}f\left(\frac{r}{2n+1}\right)=\frac{1}{1+\sqrt{3}}+199$ then the value of $n$ equals

• A. $98$
• B. $199$
• C. $198$
• D. $200$

Answer 1

Answer: (B)

$199$

Given $f\left(x\right)=\frac{3^x}{\sqrt{3}+3^x}...\left(i\right)$
$f(1-x)=\frac{\sqrt{3}}{3^x+\sqrt{3}}.....\left(ii\right)$
$\therefore f\left(x\right)+f(1-x)=1$
Now ${\sum }_{r=0}^{2n}f\left(\frac{r}{2n+1}\right)=f\left(0\right)+{\sum }_{r=1}^{2n}\left(\frac{r}{2n+1}\right)(r=0,1,2,....2n)$
$=f\left(0\right)+f\left(\frac{1}{2n+1}\right)+f\left(\frac{2}{2n+1}\right)+.....+f\left(\frac{2n-1}{2n+1}\right)+f\left(\frac{2n}{2n+1}\right)$
$=f\left(0\right)+f\left(\frac{1}{2n+1}\right)+f\left(\frac{2n}{2n+1}\right)+f\left(\frac{2}{2n+1}\right)+f\left(\frac{2n-1}{2n+1}\right)+....+upton$
${\sum }_{r=0}^{2n}f\left(\frac{r}{2n+1}\right)=\frac{1}{1+\sqrt{3}}+1+1+....+1$
$\Rightarrow \frac{1}{1+\sqrt{3}}+199=\frac{1}{1+\sqrt{3}}+n\Rightarrow n=199$