If `f(x) = (a^(x))/(a^(x) + sqrt(a)), a gt 0` prove that
(i) `f(x) + f(1-x) = 1`
(ii) `underset(k=1)overset(2n-1)sum 2f ((k)/(2n)) = 2n-1`
(iii) `underset(k=1)overset(2n)sum 2f((k)/(2n+1)) = 2n`

Explanation

(i) Given functions f(x) is `f(x) = (a^(x))/(a^(x) + sqrt(a)), a gt 0`
`f(1-x) = (a^(1-x))/(a^(1-x) + sqrt(a)) = (a)/(a+a^(x)sqrt(a)) = (sqrt(a))/(a^(x) + sqrt(a))`
`:. f(x) + f(1-x) = (a^(x))/(a^(x) + sqrt(a)) + (sqrt(a))/(a^(x) + sqrt(a)) = (a^(x) + sqrt(a))/(a^(x) + sqrt(a)) = 1`
(ii) `underset(k=1)overset(2n-1)sum 2.f ((k)/(2n)) = 2[f((1)/(2n)) + f((2)/(2n))+.....+ f((2n-2)/(2n)) + f((2n-1)/(2n))]`
`= 2[(f((1)/(2n)) + f((2n-1)/(2n)) + (f((2-1)/(2n)) + (f((2)/(2n))+f((2n-2)/(2n)))+....+(f((n-1)/(2n)) + f((n+1)/(2n)) + f((n)/(2n)))]`
`= 2[underset("(n-1) times")((1+1+...+1))+ f((1)/(2))]`
`{:(=2[(n-1)+(1)/(2)],,because f(x)+f(1-x)=1),(=2n-1,,because f((1)/(2))+f((1)/(2))=1),(,," "f((1)/(2)) = ((1)/(2))):}`
(iii) `underset(k=1)overset(2n)sum 2f((k)/(2n+1)) = 2[f((1)/(2n+1)) + f((2)/(2n+1))+....+f((2n-1)/(2n+1)) + f((2n)/(2n+1))]`
`= 2[(f((1)/(2n+1))+f((2n)/(2n+1))) + (f((2)/(2n+1))+f((2n-1)/(2n+1))) +....+(f((n)/(2n+1)) + f((n+1)/(2n+1)))]`
`= 2[underset("n times")((1+1+....+1))]`