Inspired by AMC 12A #22 (2020)

Squaring the given equality, we can find that the rational part $Q(n)$ of $(3+\sqrt{8})^n$ is equal to $c^{2}_{n} + 2 d^{2}_{n}$ .

Knowing that $3+\sqrt{8}$ is the root of the characteristic polynomial $x^2-6x+1 = 0$ , we find the recursive relation $Q(n+2) = 6Q(n+1) - Q(n)$ , for $Q(0)=1, Q(1)=3$ . The exact formula for this recursive $Q(n)$ can be shown to be

$Q(n) = \dfrac{(3+\sqrt{8})^n + (3-\sqrt{8})^n}{2}$

Notice that $\left|\frac{3+\sqrt{8}}{7}\right|, \left|\frac{3-\sqrt{8}}{7}\right| <1$ . Thus we can apply an infinite sum of a geometric progression to get

$\displaystyle \sum_{n=0}^{\infty} \dfrac{c^{2}_{n} + 2 d^{2}_{n}}{7^n} = \frac{1}{2}\left(\frac{1}{1-\frac{3+\sqrt{8}}{7}}+\frac{1}{1-\frac{3-\sqrt{8}}{7}}\right) = \frac{7}{2}\left(\frac{1}{4-\sqrt{8}}+\frac{1}{4+\sqrt{8}}\right) = \boxed{\frac{7}{2}.}$