find-the-infinite-sum-n-0-F-n-2-n-where-F-n-1-5-1-5-2-n-1-1-5-1-5-2-n-1-

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

f i n d t h e i n f i n i t e s u m \underset{n = 0}{\sum^{\infty}} \frac{F_{n}}{2^{n}}

w h e r e F_{n} = \frac{1}{\sqrt{5}} {(\frac{1 + \sqrt{5}}{2})}^{n + 1} - \frac{1}{\sqrt{5}} {(\frac{1 - \sqrt{5}}{2})}^{n + 1}

Commented by abdomathmax last updated on 25/Apr/20

w e h a v e 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}^{\infty} \frac{F_{n}}{2^{n}} = \frac{1 + \sqrt{5}}{2 \sqrt{5}} \sum_{n = 0}^{\infty} {(\frac{1 + \sqrt{5}}{4})}^{n} - \frac{1 - \sqrt{5}}{2 \sqrt{5}} \sum_{n = 0}^{\infty} {(\frac{1 - \sqrt{5}}{4})}^{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

I g o t t h e s a m e r e s u l t, 4 . T h a n k s f o r

y o u r a n s w e r .

Commented by abdomathmax last updated on 25/Apr/20

y o u a r e w e l c o m e