Geometrische Reihe/k-te Wurzel aus 2/Einheitsnachweis/Aufgabe/Lösung

Es ist

${}{\begin{aligned}&\,\,\,\,\,\,\,{\left(1-{\sqrt[{k}]{2}}\right)}\cdot {\left(1+{\sqrt[{k}]{2}}+{\sqrt[{k}]{2}}^{2}+\cdots +{\sqrt[{k}]{2}}^{k-2}+{\sqrt[{k}]{2}}^{k-1}\right)}\\&=1+{\sqrt[{k}]{2}}+{\sqrt[{k}]{2}}^{2}+\cdots +{\sqrt[{k}]{2}}^{k-2}+{\sqrt[{k}]{2}}^{k-1}-{\sqrt[{k}]{2}}{\left(1+{\sqrt[{k}]{2}}+{\sqrt[{k}]{2}}^{2}+\cdots +{\sqrt[{k}]{2}}^{k-2}+{\sqrt[{k}]{2}}^{k-1}\right)}\\&=1+{\sqrt[{k}]{2}}+{\sqrt[{k}]{2}}^{2}+\cdots +{\sqrt[{k}]{2}}^{k-2}+{\sqrt[{k}]{2}}^{k-1}-{\sqrt[{k}]{2}}-{\sqrt[{k}]{2}}^{2}-{\sqrt[{k}]{2}}^{3}-\cdots -{\sqrt[{k}]{2}}^{k-1}-{\sqrt[{k}]{2}}^{k}\\&=1-{\sqrt[{k}]{2}}^{k}\\&=1-2\\&=-1.\,\end{aligned}}$

Zur gelösten Aufgabe