Question

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

Answer 1

$f ( x )= A _0+\displaystyle\sum_{ k =1}^{20} A _{ k } x ^{ k }=\displaystyle\sum_{ k =0}^{20} A _{ k } x ^{ k }$
$\displaystyle\sum_{ r =0}^6 f \left(\alpha^{ r } x \right)= f ( x )+ f (\alpha x )+ f \left(\alpha^2 x \right)+\ldots \ldots+ f \left(\alpha^6 x \right) $
$=\displaystyle\sum_{ k =0}^{20}\left( A _{ k } x ^{ k }+ A _{ k }(\alpha x )^{ k }+ A _{ k }\left(\alpha^2 x \right)^{ k }+\ldots \ldots+ A _{ k }\left(\alpha^6 x \right)^{ k }\right.$
$\displaystyle\sum_{ k =0}^{20} A _{ k } x ^{ k }\left(1+\alpha^{ k }+\left(\alpha^2\right)^{ k }+\left(\alpha^3\right)^{ k }+\ldots \ldots .+\left(\alpha^6\right)^{ k }\right] $
$= A _0 x ^0(7)+ A _7 x ^7(7)+ A _{14} x ^{14}(7) $
$=7\left( A _0+ A _7 x ^7+ A _{14} x ^{14}\right)$
$n=7$