Question

If $f\left(x\right)=\frac{a^x}{a^x+\sqrt{a}}(a>0)$, then ${\sum }_{r=1}^{2n-1}2f\left(\frac{r}{2n}\right)$ is equal to

• A. $1$
• B. $2n$
• C. $2n-1$
• D. $\frac{(2n-1)a}{2}$

Answer 1

Answer: (C)

$2n-1$

$\begin{array}{cc}\text{Solution}& -f\left(x\right)=\frac{a^x}{a^x+\sqrt{a}}(a>0)\\ & \Rightarrow \sum _{x=1}^{2n-1}2f\left(\frac{r}{2n}\right)\\ & \Rightarrow 2f\left(\frac{1}{2n}\right)+2f\left(\frac{2}{2n}\right)+\cdots +2f\left(\frac{2n-2}{2n}\right)+2f\left(\frac{2n-1}{2n}\right)\\ & \Rightarrow 2[f\left(\frac{1}{2n}\right)+f\left(\frac{2}{2n}\right)+\cdots +f\left(\frac{n}{2n}\right)+f\left(\frac{n+1}{2n}\right)+\cdots +f\left(\frac{2n-1}{2n}\right)]\end{array}$ (1)

Now, $\begin{array}{cc}f\left(x\right)+f(1-x)& =\frac{a^x}{a^x+\sqrt{a}}+\frac{\sqrt{a}}{\sqrt{a}+a^x}\\ & =\frac{a^x+\sqrt{a}}{a^x+\sqrt{a}}=1-\left(2\right)\end{array}$

$\begin{array}{c}\text{Using (2) we get}\\ \Rightarrow 2[f\left(\frac{1}{2n}\right)+f\left(\frac{2n-1}{2n}\right)+\dots +f\left(\frac{n-1}{2n}\right)+f\left(\frac{n+1}{2n}\right)]+2f\left(\frac{n}{2n}\right)\\ \Rightarrow 2(n-1)+2f\left(\frac{1}{2}\right)-\left(3\right)\\ \text{Now put}x=\frac{1}{2}\text{in}f\left(x\right)+f(1-x)\\ \Rightarrow f\left(\frac{1}{2}\right)+f\left(\frac{1}{2}\right)=1\\ \Rightarrow f\left(\frac{1}{2}\right)=\frac{1}{2}\end{array}$

$\begin{array}{c}\text{from equation}3\\ \Rightarrow {\sum }_{n=1}^{2n-1}2f\left(\frac{n}{2n}\right)=2(n-1)+\frac{1}{2}\\ =2(n-1+\frac{1}{2})\\ =2n-1\\ \text{then},\left(c\right)\text{is the correct option.}\end{array}$