If $2f\left(xy\right)={\left(f\left(x\right)\right)}^y+{\left(f\left(y\right)\right)}^x\forall x,y\in R$  and $f\left(1\right)=a(\ne 1)$ , then ${\displaystyle \sum _{k=1}^{n}f\left(k\right)}$ =

$2f\left(xy\right)={\left(f\left(x\right)\right)}^y+{\left(f\left(y\right)\right)}^x\forall x,y\in R$ , $x,y\in R$  and $f\left(1\right)=a$ 

Putting $x=1$ , $y=2$ , we get

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

Putting $x=1$ , $y=3$ , we get

$2f\left(3\right)={\left(f\left(1\right)\right)}^3+f\left(3\right)\Rightarrow f\left(3\right)=a^3$ , and so on.

$\therefore f\left(n\right)=a^n$

${\displaystyle \sum _{i=1}^{n}f\left(i\right)=a+a^2+\mathrm{...}+a^n}=\frac{a(a^n-1)}{a-1}$ .$$