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Question Number 109460 by 150505R last updated on 23/Aug/20

Answered by mathmax by abdo last updated on 23/Aug/20

$$\mathrm{the}\mathrm{roots}\mathrm{are}{\mathrm{z}}_1=\frac{-\mathrm{b}+\mathrm{i}\sqrt{4\mathrm{ac}-{\mathrm{b}}^2}}{2\mathrm{a}}\mathrm{and}{\mathrm{z}}_2=\frac{-\mathrm{b}-\mathrm{i}\sqrt{4\mathrm{ac}-{\mathrm{b}}^2}}{2\mathrm{a}}\Rightarrow$$$$\mid {\mathrm{z}}_1\mid =\frac{1}{2\mid \mathrm{a}\mid }\sqrt{{\mathrm{b}}^2+4\mathrm{ac}-{\mathrm{b}}^2}=\frac{2\sqrt{\mathrm{ac}}}{2\mid \mathrm{a}\mid }=\sqrt{\frac{\mathrm{c}}{\mathrm{a}}}<1\mathrm{also}\mid {\mathrm{z}}_2\mid =\sqrt{\frac{\mathrm{c}}{\mathrm{a}}}\Rightarrow$$$$\mathrm{f}\left(\mathrm{n}\right)=\sum _{\mathrm{k}=\mathrm{o}}^{\mathrm{n}}{\left(\frac{\mathrm{c}}{\mathrm{a}}\right)}^{\frac{\mathrm{k}}{2}}+{\left(\frac{\mathrm{c}}{\mathrm{a}}\right)}^{\frac{\mathrm{k}}{2}}-2=2\sum _{\mathrm{k}=0}^{\mathrm{n}}{\left(\sqrt{\frac{\mathrm{c}}{\mathrm{a}}}\right)}^{\mathrm{k}}-2=2\frac{1-{\left(\sqrt{\frac{\mathrm{c}}{\mathrm{a}}}\right)}^{\mathrm{n}+1}}{1-\sqrt{\frac{\mathrm{c}}{\mathrm{a}}}}-2$$$$\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{n}\right)=\frac{2}{1-\frac{\sqrt{\mathrm{c}}}{\sqrt{\mathrm{a}}}}-2=\frac{2\sqrt{\mathrm{a}}}{\sqrt{\mathrm{a}}-\sqrt{\mathrm{c}}}-2=\frac{2\sqrt{\mathrm{a}}-2\sqrt{\mathrm{a}}+2\sqrt{\mathrm{c}}}{\sqrt{\mathrm{a}}-\sqrt{\mathrm{c}}}$$$$=\frac{2\sqrt{\mathrm{c}}}{\sqrt{\mathrm{a}}-\sqrt{\mathrm{c}}}\mathrm{the}\mathrm{true}\mathrm{answer}\mathrm{is}\left(\mathrm{d}\right)$$

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