Question-89852

Question Number 89852 by M±th+et£s last updated on 19/Apr/20

Answered by mr W last updated on 19/Apr/20

F_{n} = \frac{1}{\sqrt{5}} (φ^{n} - ψ^{n}) w i t h

φ = \frac{1 + \sqrt{5}}{2}, ψ = \frac{1 - \sqrt{5}}{2} = 1 - φ = - \frac{1}{φ}

x^{n} F_{n} = \frac{1}{\sqrt{5}} [{(x φ)}^{n} - {(x ψ)}^{n}]

\underset{n = 1}{\sum^{\infty}} x^{n} F_{n} = \frac{1}{\sqrt{5}} [\underset{n = 1}{\sum^{\infty}} {(x φ)}^{n} - \underset{n = 1}{\sum^{\infty}} {(x ψ)}^{n}]

= \frac{1}{\sqrt{5}} [\frac{x φ}{1 - x φ} - \frac{x ψ}{1 - x ψ}]

= \frac{x}{\sqrt{5}} [\frac{φ}{1 - x φ} + \frac{1}{φ + x}] = 1

\Rightarrow \frac{φ}{\frac{1}{x} - φ} + \frac{1}{\frac{φ}{x} + 1} = \sqrt{5}

l e t t = \frac{1}{x}

\Rightarrow \frac{φ}{t - φ} + \frac{1}{φ t + 1} = \sqrt{5}

t^{2} + \frac{\sqrt{5} (1 - φ^{2}) - 1 - φ^{2}}{\sqrt{5} φ} t - 1 = 0

t^{2} - \frac{(\sqrt{5} + 1) φ + 2}{\sqrt{5} φ} t - 1 = 0

t^{2} - 2 t - 1 = 0

\Rightarrow t = 1 + \sqrt{2} (x > 0)

\Rightarrow x = \frac{1}{t} = \frac{1}{1 + \sqrt{2}} = \sqrt{2} - 1

Commented by M±th+et£s last updated on 19/Apr/20

n i c e w o r k t h a n k y o u s i r

Commented by mr W last updated on 19/Apr/20

a n s w e r c o r r e c t ?

i d i d i t i n h u r r y, n o t v e r y c a r e f u l l y

a n d n o t c h e c k e d .

Commented by M±th+et£s last updated on 19/Apr/20

y e s s i r i t s c o r r e c t