<s>[INST] How to simplify expression with Fibonacci numbers I have to simplify the expression $\sum_{n=0}^\infty \sum_{k=0}^n \frac{F_{2k}F_{n-k}}{10^n}$. I only noticed that $\sum_{n=0}^\infty \sum_{k=0}^n \frac{F_{2k}F_{n-k}}{10^n} = \sum_{n=0}^\infty  \frac{1}{10^n} \sum_{k=0}^n F_{2k}F_{n-k}$. What to do next?
 [/INST] The generating function of the Fibonacci numbers is well-known:
$$ \frac{x}{1-x-x^2} = \sum_{n=0}^\infty F_n x^n. $$
We also want to isolate only the even Fibonacci numbers:
$$ \frac{x}{1-x-x^2} + \frac{-x}{1+x-x^2} = \sum_{n=0}^\infty F_n (x^n + (-x)^n) = \sum_{n=0}^\infty 2F_{2n} x^{2n}. $$
Since
$$
\frac{x}{1-x-x^2} + \frac{-x}{1+x-x^2} =
\frac{x(1+x-x^2) -x(1-x-x^2)}{(1-x-x^2)(1+x-x^2)} =
\frac{2x^2}{(1-x^2)^2-x^2},
$$
we can conclude that
$$
\frac{x}{1-3x+x^2} = \sum_{n=0}^\infty F_{2n} x^n.
$$
It follows that
$$
\frac{x}{1-x-x^2} \frac{x}{1-3x+x^2} = \sum_{n=0}^\infty x^n \sum_{k=0}^n F_{2k} F_{n-k}.
$$
In particular, your expression equals $100/6319$.
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