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Question Number 148501 by mathmax by abdo last updated on 28/Jul/21

$$\mathrm{let}{\mathrm{U}}_{\mathrm{n}}=\{\mathrm{z}\in \mathrm{C}/{\mathrm{z}}^{\mathrm{n}}=1\}\mathrm{simplify}$$$$\sum _{\mathrm{p}=0}^{2\mathrm{n}-1}{\mathrm{w}}^{\mathrm{p}}\mathrm{with}\mathrm{w}\in {\mathrm{U}}_{\mathrm{n}}$$$$\mathrm{and}\sum _{\mathrm{p}=0}^{2\mathrm{n}-1}{(2\mathrm{w}+1)}^{\mathrm{p}}$$

Answered by Olaf_Thorendsen last updated on 28/Jul/21

$$\bullet w=1$$$${\mathrm{S}}_n\left(1\right)=\underset{p=0}{\sum ^{2n-1}}{1}^p=2n\left(\mathrm{trivial}\right)$$$$$$$$\bullet w^n=1\mathrm{and}w\ne 1$$$${\mathrm{S}}_n\left(w\right)=\underset{p=0}{\sum ^{2n-1}}{w}^p=\frac{1-w^{2n}}{1-w}=\frac{1-1}{1-w}=0$$$$$$$${\mathrm{T}}_n\left(w\right)=\underset{p=0}{\sum ^{2n-1}}{(2w+1)}^p$$$${\mathrm{T}}_n\left(w\right)=\frac{1-{(2w+1)}^{2n}}{1-(2w+1)}=\frac{{(2w+1)}^{2n}-1}{2w}$$

Answered by mathmax by abdo last updated on 28/Jul/21

$$\mathrm{if}\mathrm{w}=1{\mathrm{A}}_{\mathrm{n}}=\sum _{\mathrm{p}=0}^{2\mathrm{n}-1}{\mathrm{w}}^{\mathrm{p}}=2\mathrm{n}$$$$\mathrm{if}\mathrm{w}\ne 1\sum _{\mathrm{p}=0}^{2\mathrm{n}-1}{\mathrm{w}}^{\mathrm{p}}=\frac{1-{\mathrm{w}}^{2\mathrm{n}}}{1-\mathrm{w}}=\frac{1-{\left({\mathrm{w}}^{\mathrm{n}}\right)}^2}{1-\mathrm{w}}=\frac{1-1}{1-\mathrm{w}}=0$$$${\mathrm{alsoB}}_{\mathrm{n}}=\sum _{\mathrm{p}=0}^{2\mathrm{n}-1}{(2\mathrm{w}+1)}^{\mathrm{p}}\mathrm{the}\mathrm{roots}\mathrm{of}{\mathrm{z}}^{\mathrm{n}}=1\mathrm{are}{\mathrm{z}}_{\mathrm{k}}={\mathrm{e}}^{\mathrm{i}\frac{2\mathrm{k}\pi }{\mathrm{n}}}$$$$\mathrm{and}0⩽\mathrm{k}⩽\mathrm{n}-1\Rightarrow \mathrm{w}\ne -\frac{1}{2}\Rightarrow {\mathrm{B}}_{\mathrm{n}}=\frac{1-{(2\mathrm{w}+1)}^{2\mathrm{n}}}{1-(2\mathrm{w}+1)}$$$$=\frac{1}{2\mathrm{w}}\{{(2\mathrm{w}+1)}^{2\mathrm{n}}-1\}$$$$=\frac{1}{2\mathrm{w}}\{\sum _{\mathrm{k}=0}^{2\mathrm{n}}{\mathrm{C}}_{2\mathrm{n}}^{\mathrm{k}}{\left(2\mathrm{w}\right)}^{\mathrm{k}}-1\}$$$$=\frac{1}{2\mathrm{w}}\sum _{\mathrm{k}=0}^{2\mathrm{n}}{\mathrm{C}}_{2\mathrm{n}}^{\mathrm{k}}{2}^{\mathrm{k}}{\mathrm{w}}^{\mathrm{k}}$$$$$$

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