Question

If $\alpha =e^{i\frac{2\pi }{7}}$ and $f(x)=A_0+\sum _{k=1}^{20}{A}_k{x}^k$, then the value of $\sum _{r=0}^{6}f\left({\alpha }^{r}x\right)=n\left(A_0+A_n{x}^n+A_{2n}{x}^{2n}\right)$, then find the value of ' $n$ '

Answer 1

Answer: 7

$f(x)=A_0+\sum _{k=1}^{20}{A}_k{x}^k=\sum _{k=0}^{20}{A}_k{x}^k$
$\sum _{r=0}^{6}f\left({\alpha }^{r}x\right)=f(x)+f(\alpha x)+f\left({\alpha }^{2}x\right)+\dots \dots +f\left({\alpha }^{6}x\right)$
$=\sum _{k=0}^{20}(A_k{x}^k+A_k(\alpha x{)}^k+A_k{\left({\alpha }^{2}x\right)}^k+\dots \dots +A_k{\left({\alpha }^{6}x\right)}^k$
$\sum _{k=0}^{20}{A}_k{x}^k\left(1+{\alpha }^k+{\left({\alpha }^2\right)}^k+{\left({\alpha }^3\right)}^k+\dots \dots .+{\left({\alpha }^6\right)}^k\right]$
$=A_0{x}^0(7)+A_7{x}^7(7)+A_{14}{x}^{14}(7)$
$=7\left(A_0+A_7{x}^7+A_{14}{x}^{14}\right)$
$n=7$