If `f(x)=(a^x)/(a^x+sqrt(a ,)),(a >0),` then find the value of g(n)=`sum_(r=1)^(2n-1)2f(r/(2n))` g(4)

Explanation

`f(x)=(a^(x))/(a^(x)+sqrt(a))`
or `f(1-x)=(a^(1-x))/(a^(1-x)+sqrt(a))`
`=(a^(1))/(a^(1)+sqrt(a)a^(x))`
`=(sqrt(a))/(sqrt(a)+a^(x))`
` or f(x)+f(1-x)=1`
` :. f((1)/(2))=(1)/(2)`
so, `sum_(r=1)^(2n-1)2f((r)/(2n))=2[f((1)/(2n))+f((2)/(2n))+ ... +f((n-1)/(2n))+f((n)/(2n))+f((n+1)/(2n))+ ... +f((2n-1)/(2n))]`
`=2{[f((1)/(2n))+f((2n-1)/(2n))]+[f((2)/(2n))+f((2n-2)/(2n))]+ ... +[f((n-1)/(2n))+f((n+1)/(2n))]+f((1)/(2))}`
`=2{[f((1)/(2n))+f(1-(1)/(2n))]+[f((2)/(2n))+f(1-(2)/(2n))]+ ... +[f((n-1)/(2n))+f(1-(n-1)/(2n))]+(1)/(2)}`
`=2[1+1+1+...+(n-1)" times "]+1`
`=2n-1`