Summation involving Fibonacci numbers

Question

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.

Answer 1

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

Answer 2

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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