3
$\begingroup$

Find: $$ \sum_{n=0}^\infty \sum_{k=0}^n \frac{F_{2k}F_{n-k}}{10^n} $$ where $F_n$ is $n$-th Fibonacci number.

$\endgroup$

closed as off-topic by heropup, iadvd, user91500, Claude Leibovici, Watson Sep 13 '16 at 9:12

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, iadvd, user91500, Claude Leibovici, Watson
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ Have you used generating functions before? If so, you just need to find the ordinary generating functions for $F_n$ and $F_{2n}$. Then multiply them and plug in $x = 1/10$. $\endgroup$ – André 3000 Mar 31 '14 at 19:25
  • 1
    $\begingroup$ This has a basic answer at this MSE link. $\endgroup$ – Marko Riedel Aug 5 '14 at 22:18
3
$\begingroup$

We have: $$ \sum_{k=0}^{+\infty}F_{2k}x^k = \frac{x}{1-3x+x^2} \tag{1}$$ and $$ \sum_{k=0}^{+\infty}F_{k} x^k = \frac{x}{1-x-x^2}\tag{2} $$ hence: $$ \sum_{n\geq 0}\frac{1}{10^n}\sum_{k=0}^{n}F_{2k}F_{n-k}=\left.\frac{x^2}{(1-3x+x^2)(1-x-x^2)}\right|_{x=\frac{1}{10}} = \color{red}{\frac{100}{6319}}.\tag{3}$$

$\endgroup$
2
$\begingroup$

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{% \sum_{n = 0}^{\infty}\,\sum_{k = 0}^{n}{F_{2k}\, F_{n - k} \over 10^{n}}} & = \sum_{k = 0}^{\infty}F_{2k}\, \sum_{n = k}^{\infty}{F_{n - k} \over 10^{n}} = \sum_{k = 0}^{\infty}{F_{2k} \over 10^{k}} \sum_{n = 0}^{\infty}{F_{n} \over 10^{n}} \\[5mm] & = \bracks{\sum_{n = 0}^{\infty}F_{n}\pars{1 \over 10}^{n}} \bracks{\half\sum_{k = 0}^{\infty}{F_{k}\pars{1 \over\root{10}}^{k}} + \half\sum_{k = 0}^{\infty}{F_{k} \pars{-\,{1 \over \root{10}}}^{k}}} \end{align}


With the Fibonacci generating function $\ds{\,\mc{F}\pars{z} = \sum_{n = 0}^{\infty}F_{n}\,z^{n} = {z \over 1 - z - z^{2}}}$: \begin{align} \color{#f00}{% \sum_{n = 0}^{\infty}\,\sum_{k = 0}^{n}{F_{2k}\, F_{n - k} \over 10^{n}}} & = \half\,\mc{F}\pars{1 \over 10}\bracks{\mc{F}\pars{1 \over \root{10}} + \mc{F}\pars{-\,{1 \over \root{10}}}} = \color{#f00}{100 \over 6319} \approx 0.0158 \end{align}

Note that $\ds{\quad\mc{F}\pars{1 \over 10} = {10 \over 89}\quad \mbox{and}\quad \,\mc{F}\pars{\pm\,{1 \over \root{10}}} = {10 \pm 9\root{10} \over 71}}$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.