calculate-n-1-F-n-n-3-n-

Question Number 170580 by mnjuly1970 last updated on 27/May/22

c a l c u l a t e

\underset{n = 1}{\sum^{\infty}} \frac{F_{n}}{n . 3^{n}} = ?

Answered by Mathspace last updated on 27/May/22

f_{n} i s f e f i m e d b y f_{0} = 0 a n d ? f_{1} = 1

a n d f_{n + 2} = f_{n} + f_{n + 1}

c e \to r^{2} - r - 1 = 0

Δ = 1 + 4 = 5 \Rightarrow r_{1} = \frac{1 + \sqrt{5}}{2}

r_{2} = \frac{1 - \sqrt{5}}{2}

\Rightarrow f_{n} = {a r}_{1}^{n} + {b r}_{2}^{n}

f_{0} = 0 = a + b

f_{1} = 1 = r_{1} a + r_{2} b = (r_{1} - r_{2}) a

= \sqrt{5} a \Rightarrow a = \frac{1}{\sqrt{5}} a n d b = - \frac{1}{\sqrt{5}}

\Rightarrow f_{n} = \frac{1}{\sqrt{5}} r_{1}^{n} - \frac{1}{\sqrt{5}} r_{2}^{n} \Rightarrow

\sum_{n = 1}^{\infty} \frac{f_{n}}{n 3^{n}} = \frac{1}{\sqrt{5}} \sum_{n = 1}^{\infty} \frac{1}{n} {(\frac{r_{1}}{3})}^{n}

- \frac{1}{\sqrt{5}} \sum_{n = 1}^{\infty} \frac{1}{n} {(\frac{r_{2}}{3})}^{n}

l e t s (x) = \sum_{n = 1}^{\infty} \frac{x^{n}}{n} \Rightarrow

s (x) = - l n (1 - x) \Rightarrow

\sum_{n = 1}^{\infty} \frac{f_{n}}{n 3^{n}} = - \frac{1}{\sqrt{5}} l n (1 - \frac{r_{1}}{3})

+ \frac{1}{\sqrt{5}} l n (1 - \frac{r_{2}}{3})

= \frac{1}{\sqrt{5}} l n (\frac{3 - r_{2}}{3 - r_{1}})

= \frac{1}{\sqrt{5}} l n (\frac{3 - \frac{1 - \sqrt{5}}{2}}{3 - \frac{1 + \sqrt{5}}{2}})

= \frac{1}{\sqrt{5}} l n (\frac{5 + \sqrt{5}}{5 - \sqrt{5}}) = \frac{1}{\sqrt{5}} l n (\frac{\sqrt{5} + 1}{\sqrt{5} - 1})

0

Commented by mnjuly1970 last updated on 28/May/22

t h a n k s a l o t \dots

Commented by Tawa11 last updated on 08/Oct/22

Great sir

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