F-0-0-F-1-1-F-n-1-F-n-F-n-1-prove-1-89-i-1-10-i-F-i-1-

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

F (0) = 0 F (1) = 1 F (n + 1) = F (n) + F (n - 1)

p r o v e :

\frac{1}{89} = \underset{i = 1}{\sum^{+ \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 (n) = A {(\frac{1 + \sqrt{5}}{2})}^{n} + B {(\frac{1 - \sqrt{5}}{2})}^{n}

F (0) = A + B = 0 \Rightarrow B = - A

F (1) = A (\frac{1 + \sqrt{5}}{2}) - A (\frac{1 - \sqrt{5}}{2}) = 1 \Rightarrow A = \frac{1}{\sqrt{5}}

\Rightarrow F (n) = \frac{1}{\sqrt{5}} [{(\frac{1 + \sqrt{5}}{2})}^{n} - {(\frac{1 - \sqrt{5}}{2})}^{n}]

\underset{n = 1}{\sum^{\infty}} 10^{- n} F (n - 1)

= \underset{n = 0}{\sum^{\infty}} 10^{- (n + 1)} F (n)

= \frac{1}{10 \sqrt{5}} \underset{n = 0}{\sum^{\infty}} \frac{1}{10^{n}} [{(\frac{1 + \sqrt{5}}{2})}^{n} - {(\frac{1 - \sqrt{5}}{2})}^{n}]

= \frac{1}{10 \sqrt{5}} \underset{n = 0}{\sum^{\infty}} [{(\frac{1 + \sqrt{5}}{20})}^{n} - {(\frac{1 - \sqrt{5}}{20})}^{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}} (\frac{2 \sqrt{5}}{356})

= \frac{1}{89} ✓

Commented by liuxinnan last updated on 09/Sep/24

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