Summation involving Fibonacci numbers 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.
 A: 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:
$$ \sum_{n\geq 0}\frac{1}{10^n}\sum_{k=0}^{n}F_{2k}F_{n-k}=\left.\frac{x^2}{(1-3x+x^2)(1-x-x^2)}\right|_{x=\frac{1}{10}} = \color{red}{\frac{100}{6319}}.\tag{3}$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\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 $\ds{\quad\mc{F}\pars{1 \over 10} = {10 \over 89}\quad
\mbox{and}\quad
\,\mc{F}\pars{\pm\,{1 \over \root{10}}} = {10 \pm 9\root{10} \over 71}}$.

