Question

$f\left(x\right)$ is a polynomial such that $f\left(x\right)\cdot f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right)$ such that $f\left(3\right)=28$. Then the value of $\sum _{k=1}^{10}f\left(k\right)$ is

Answer 1

$f\left(x\right)\cdot f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right)$
$\Rightarrow f\left(x\right)=1\pm x^n$
$f\left(3\right)=28\Rightarrow 1\pm 3^n=28$ $\Rightarrow \pm 3^n=27\Rightarrow n=3$
(Consider positive sign)
$\therefore f\left(x\right)=1+x^3$
$\text{Now}\sum _{k=1}^{10}f\left(k\right)$ $=\sum _{k=1}^{10}(1+k^3)$
$=10+\sum _{k=1}^{10}{k}^3=10+{\left(\frac{10\times 11}{2}\right)}^2=3035$