A Lucas Sum II

The Binet formula $F_{n} = \frac{\phi^{n}-(-\phi)^{-n}}{\sqrt{5}}$ where $\phi = \frac{1+\sqrt{5}}{2}$ is well-known, but not so much the analogous formula for the Lucas numbers, which is key to this problem. It turns out that

$L_{n} \equiv \phi^{n}+(-\phi)^{-n}$

for all $n \geq 0$ . So we rewrite the sum as

$\sum_{n=0}^{\infty} \frac{L_{n}^{2}}{4^{n}} = \sum_{n=0}^{\infty} \frac{(\phi^{n}+(-\phi)^{-n})^{2}}{4^{n}}$

$= \sum_{n=0}^{\infty} \left( \frac{\phi^{2}}{4} \right)^{n}+2 \left( -\frac{1}{4} \right)^{n}+\left( \frac{-\phi^{-2}}{4} \right)^{n}$

We split this up into three sums, each a convergent geometric series. We then get, after a bit of simplification with the help of the difference of two squares,

$\frac{1}{1-\phi^{2}/4}+\frac{2}{1-(-1/4)}+\frac{1}{1-(-\phi^{-2}/4)} = \frac{8}{5-\sqrt{5}}+\frac{8}{5}+\frac{8}{5+\sqrt{5}}$

$= \frac{8(5+\sqrt{5}+5-\sqrt{5})}{5^{2}-5}+\frac{8}{5} = \boxed{\frac{28}{5}}$