If $\alpha=e^{2 \pi i / 7}$ and $f(x)=A_{0}+\sum_{k=1}^{20} A_{k} x^{k} \\$, then find the value of, $f(x)+f(\alpha x)+\ldots+f\left(\alpha^{6} x\right)$ independent of $\alpha .$

Solution:

Given that, $ \alpha=\mathrm{e}^{2 \pi / 7} \\ $ and, $ f(\mathrm{x})=\mathrm{A}_{0}+\sum_{k=1}^{20} A_{k} x^{k} \\ $

Then, $ f\left(\alpha^{n} x\right)=\mathrm{A}_{0}+\sum_{k=1}^{6} A_{k} \alpha^{n k} x^{k} \\ $

$ \therefore \quad \mathrm{S}=f(x)+f(\alpha \mathrm{x})+\ldots \ldots . .+f\left(\alpha^{6} \mathrm{x}\right)=\sum_{k=1}^{6} f\left(\alpha^{n} x\right) \\ $

$ =\sum_{k=1}^{6}\left\{A_{0}+\sum_{k=1}^{20} A_{k} \alpha^{n k} x^{k}\right\} \\ $

$ =7 A_{0}+\sum_{k=1}^{20}\left\{A_{k} x^{k} \sum_{n=0}^{6} \alpha^{n k}\right\} \\ $

$ 1+\alpha^{\mathbf{k}}+\alpha^{2 \mathbf{k}}+\ldots \ldots \ldots+\mathrm{a}^{\mathbf{k}}=7 \\ $

$ \quad$ If $\mathrm{a}^{k}=1 \quad \\ $ i.e. $k$ is a multiple of 7. $ \quad$ If $\mathrm{a}^{k} \neq 1 \quad \ $ i.e. $k $ is not a multiple of 7. $ \therefore \quad S=7 A_{0}+7 A_{7} x^{7}+7 A_{14} x^{14}=7\left(A_{7} x^{7}+A_{14} x^{14}\right) \ $