Question

If $f\left(x\right)=\frac{a^x}{a^x+\sqrt{a}}(a>0)$,evaluate $\sum _{r=1}^{2n-1}2f\left(\frac{r}{2n}\right)$. for n=8

Answer 1

Given,
$f\left(x\right)=\frac{a^x}{a^x+\sqrt{x}}$ .....(i)
Now, $f(1-x)=\frac{a^{1-x}}{a^{1-x}+\sqrt{a}}=\frac{\sqrt{a}}{\sqrt{a}+a^x}$ ...(ii)
From Eqs. (i) and (ii); we have $f\left(x\right)=f(1-x)=1$ ...(iii)
$\Rightarrow f\left(\frac{r}{2n}\right)+f\left(\frac{2n-r}{2n}\right)=1$
$\Rightarrow \sum _{r=1}^{2n-1}f\left(\frac{r}{2n}\right)+\sum _{r=1}^{2n-1}f\left(\frac{2n-r}{2n}\right)=2n-1$
$\Rightarrow \sum _{r=1}^{2n-1}f\left(\frac{r}{2n}\right)+\sum _{t=1}^{2n-1}f\left(\frac{t}{2n}\right)=2n-1$ (putting $2n-r=t$)
Hence, $2\sum _{r=1}^{2n-1}f\left(\frac{r}{2n}\right)=2n-1=2\left(8\right)-1=15$.............. at $n=8$