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Question Number 90609 by Maclaurin Stickker last updated on 25/Apr/20

$$findtheinfinitesum\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{F_n}{2^n}$$$$where{F}_n=\frac{1}{\sqrt{5}}{\left(\frac{1+\sqrt{5}}{2}\right)}^{n+1}-\frac{1}{\sqrt{5}}{\left(\frac{1-\sqrt{5}}{2}\right)}^{n+1}$$

Commented by abdomathmax last updated on 25/Apr/20

$$wehave{F}_n=\frac{1}{2\sqrt{5}}\times \frac{{(1+\sqrt{5})}^n}{2^n}-\frac{1}{2\sqrt{5}}\frac{{(1-\sqrt{5})}^n}{2^n}\Rightarrow$$$$\sum _{n=0}^{\mathrm{\infty }}\frac{F_n}{2^n}=\frac{1+\sqrt{5}}{2\sqrt{5}}\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1+\sqrt{5}}{4}\right)}^n-\frac{1-\sqrt{5}}{2\sqrt{5}}\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1-\sqrt{5}}{4}\right)}^n$$$$=\frac{1+\sqrt{5}}{2\sqrt{5}}\times \frac{1}{1-\frac{1+\sqrt{5}}{4}}-\frac{1-\sqrt{5}}{2\sqrt{5}}\times \frac{1}{1-\frac{1-\sqrt{5}}{4}}$$$$=\frac{2}{\sqrt{5}}\times \frac{1+\sqrt{5}}{3-\sqrt{5}}-\frac{2}{\sqrt{5}}\times \frac{1-\sqrt{5}}{3+\sqrt{5}}$$$$=\frac{2}{\sqrt{5}}(\frac{1+\sqrt{5}}{3-\sqrt{5}}-\frac{1-\sqrt{5}}{3+\sqrt{5}})$$$$$$

Commented by Maclaurin Sticcker last updated on 25/Apr/20

$$Igotthesameresult,4.Thanksfor$$$$youranswer.$$

Commented by abdomathmax last updated on 25/Apr/20

$$youarewelcome$$

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