Question

Consider a real valued function $f\left(x\right)$ satisfying $2f\left(xy\right)=\left(f\right(x){)}^y+(f\left(y\right){)}^x\forall x,y\in R$ and $f\left(1\right)=p$ where $p\ne 1$ then find

$(p-1)\sum _{r=1}^{n}f\left(r\right)$.

• A. $p^{n-1}-p$
• B. $p^{n+1}$
• C. $p^{n+1}-p$
• D. $p^{n-1}$

Answer 1

The correct option is C $p^{n+1}-p$

$2f\left(xy\right)=\left(f\right(x){)}^y+(f\left(y\right){)}^x\forall x,y\in R$

Put $y=1$

$\Rightarrow 2f\left(x\right)=f\left(x\right)+\left(f\right(1){)}^x$

$\Rightarrow f\left(x\right)=p^x(\because f(1)=p)$

$\therefore \sum _{r=1}^{n}f\left(r\right)=\sum _{r=1}^n{p}^r=\frac{p^{n+1}-p}{p-1}$

$\Rightarrow (p-1)\sum _{r=1}^{n}f\left(r\right)=p^{n+1}-p$