Question

If $2f\left(xy\right)={\left(f\right(x\left)\right)}^y+{\left(f\right(y\left)\right)}^x\forall x,y\in R$ and $f\left(1\right)=3,$ then the value of $\sum _{r=1}^{10}f\left(r\right)$ is equal to

• A. $\frac{3}{2}(3^{10}-1)$
• B. $\frac{3}{2}(3^9-1)$
• C. $\frac{1}{2}(3^{10}-1)$
• D. $\frac{1}{2}(3^9-1)$

Answer 1

The correct option is A $\frac{3}{2}(3^{10}-1)$

Given, $2f\left(xy\right)={\left(f\right(x\left)\right)}^y+{\left(f\right(y\left)\right)}^x$ and $f\left(1\right)=3$
Put $y=1$
$\Rightarrow 2f\left(x\right)=f\left(x\right)+{\left(f\right(1\left)\right)}^x$
$\Rightarrow f\left(x\right)=3^x$

$\therefore \sum _{r=1}^{10}f\left(r\right)$
$=\sum _{r=1}^{10}{3}^r=3+3^2+3^3+\cdots +3^{10}=\frac{3}{2}(3^{10}-1)$
$(\because$ For G.P., $S_n=\frac{a(r^n-1)}{r-1})$