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Question Number 194020 by mnjuly1970 last updated on 26/Jun/23

F_{n} = F_{n}_{- 1} + F_{n - 2} F_{2} = F_{1} = 1

F_{n} : 1, 1, 2, 3, 5 \dots

f (x) = \underset{n = 1}{\sum^{\infty}} F_{n} x^{n} = x + x^{2} + \underset{n = 3}{\sum^{\infty}} (F_{n - 1} + F_{n - 2}) x^{n}

= x + x^{2} + \underset{n = 3}{\sum^{\infty}} F_{n - 1} x^{n} + x^{2} f (x)

= x + x^{2} + x^{2} f (x) + x \underset{n = 2}{\sum^{\infty}} F_{n} x^{n}

= x + x^{2} + x^{2} f (x) - x^{2} + x f (x)

∴ f (x) = \frac{x}{1 - x - x^{2}} (g e n e r a t i n g f u n c t i o n)

\frac{x}{1 - x - x^{2}} = \underset{n = 1}{\sum^{\infty}} F_{n} x^{n} \Rightarrow \frac{x^{2}}{1 - x - x^{2}} = \underset{n = 1}{\sum^{\infty}} F_{n} x^{n + 1}

x = \frac{1}{10} \Rightarrow \frac{\frac{1}{100}}{1 - \frac{1}{10} - \frac{1}{100}} = \underset{n = 1}{\sum^{\infty}} \frac{F_{n}}{10^{n + 1}}

\Rightarrow {\underset{n = 1}{\sum^{\infty}} \frac{F_{n}}{10^{n + 1}} = \frac{1}{89}}