Consider a real-valued function f(x) satisfying `2f(x y)=(f(x))^y+(f(y))^xAAx , y in Ra n df(1)=a ,w h e r ea!=1.` Prove that `(a-1)` `sum_(i=1)^nf(i)=a^(n+1)-a`

Explanation

We have `2f(xy)=(f(x))^(y)+(f(y))^(x)`.
Replacing y by 1, we get
`2f(x)=f(x)+(f(1))^(x) " or " f(x)=a^(x)`
or `sum_(i=1)^(n)f(i)=a+a^(2)+ ...+a^(n)=(a^(n+1)-a)/(a-1)`
or `(a-1)sum_(i=1)^(n)f(i)=a^(n+1)-a`