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Question Number 211370 by liuxinnan last updated on 07/Sep/24

$$F\left(0\right)=0F\left(1\right)=1F(n+1)=F\left(n\right)+F(n-1)$$$$prove:$$$$\frac{1}{89}=\underset{i=1}{\sum ^{+\mathrm{\infty }}}{10}^{-i}F(i-1)$$

Answered by mr W last updated on 07/Sep/24

$$p^2-p-1=0$$$$\Rightarrow p=\frac{1\pm \sqrt{5}}{2}$$$$\Rightarrow F\left(n\right)=A{\left(\frac{1+\sqrt{5}}{2}\right)}^n+B{\left(\frac{1-\sqrt{5}}{2}\right)}^n$$$$F\left(0\right)=A+B=0\Rightarrow B=-A$$$$F\left(1\right)=A\left(\frac{1+\sqrt{5}}{2}\right)-A\left(\frac{1-\sqrt{5}}{2}\right)=1\Rightarrow A=\frac{1}{\sqrt{5}}$$$$\Rightarrow F\left(n\right)=\frac{1}{\sqrt{5}}[{\left(\frac{1+\sqrt{5}}{2}\right)}^n-{\left(\frac{1-\sqrt{5}}{2}\right)}^n]$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{10}^{-n}F(n-1)$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}{10}^{-(n+1)}F\left(n\right)$$$$=\frac{1}{10\sqrt{5}}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{10}^n}[{\left(\frac{1+\sqrt{5}}{2}\right)}^n-{\left(\frac{1-\sqrt{5}}{2}\right)}^n]$$$$=\frac{1}{10\sqrt{5}}\underset{n=0}{\sum ^{\mathrm{\infty }}}[{\left(\frac{1+\sqrt{5}}{20}\right)}^n-{\left(\frac{1-\sqrt{5}}{20}\right)}^n]$$$$=\frac{1}{10\sqrt{5}}(\frac{1}{1-\frac{1+\sqrt{5}}{20}}-\frac{1}{1-\frac{1-\sqrt{5}}{20}})$$$$=\frac{2}{\sqrt{5}}(\frac{19+\sqrt{5}}{356}-\frac{19-\sqrt{5}}{356})$$$$=\frac{2}{\sqrt{5}}\left(\frac{2\sqrt{5}}{356}\right)$$$$=\frac{1}{89}✓$$

Commented by liuxinnan last updated on 09/Sep/24

$$thankssir$$

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