<s>[INST] Generating functions Find 
$$ \sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{F_{2k}F_{n-k}}{10^n}, $$
where $F_n$ - $n-th$ Fibonacci number 
Any hint how to start with this?
I have thought about convolution of two Fibonacci sequences, but there is $F_{2k}$, instead of $F_{k}$.
 [/INST] Consider $$\left(\sum_{k=0}^{\infty} F_k x^k \right) \cdot \left(\sum_{k=0}^{\infty} F_{2k} x^k \right) = \sum_{k=0}^{\infty} C_n x^n$$
Now $C_n = \displaystyle \sum_{k=0}^n F_{2k} F_{n-k}$. We now have that
\begin{align}
\sum_{k=0}^{\infty} F_k x^k & = \dfrac{x}{1-x-x^2}\\
\sum_{k=0}^{\infty} F_{2k} x^k & = \dfrac1{\phi-\psi} \left(\sum_{k=0}^{\infty} \phi^{2k} x^k - \sum_{k=0}^{\infty} \psi^{2k} x^k\right) = \dfrac1{\phi - \psi}\left(\dfrac1{1-\phi^2x} - \dfrac1{1-\psi^2x}\right)\\
& = \dfrac{(\phi + \psi)x}{1-(\phi^2+\psi^2)x + \phi^2 \psi^2 x^2} = \dfrac{x}{1-3x+x^2}
\end{align}
Hence,
$$\sum_{k=0}^{\infty} C_n x^n = \dfrac{x}{1-x-x^2} \cdot \dfrac{x}{1-3x+x^2}$$
Now plug in $x=\dfrac1{10}$ and get the answer.
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