Proof?(functions)

Consider a real valued function $\displaystyle f(x)$ satisfying, $\displaystyle 2f(xy) = (f(x))^y + (f(y))^x$ for all real $\displaystyle x\ \mbox{\&}\ y$ and $\displaystyle f(1) = a$ where $\displaystyle a\neq 1$. Prove that $\displaystyle (a - 1)\sum_{i = 1}^{n}f(i) = a^{n + 1} - a$.

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Originally Posted by fardeen_gen

Consider a real valued function $\displaystyle f(x)$ satisfying, $\displaystyle 2f(xy) = (f(x))^y + (f(y))^x$ for all real $\displaystyle x\ \mbox{\&}\ y$ and $\displaystyle f(1) = a$ where $\displaystyle a\neq 1$. Prove that $\displaystyle (a - 1)\sum_{i = 1}^{n}f(i) = a^{n + 1} - a$.

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Taking y= 1, $\displaystyle 2f(x)= f(x)+ f(1)^x= f(x)+ a^x$. That is, $\displaystyle f(x)= a^x$. That should be all you need.