TITLE: $a_n=20a_{n-1}+12a_{n-2}$ recurrence relation question
QUESTION [5 upvotes]: Respected Sir,
Please solve the below problem. Please...

Consider the infinite $\displaystyle\mathbb{S}=\sum_{n=0}^{\infty}\frac{a_n}{10^{2n}}$, where the sequence $\{a_n\}$ is defined by $a_0=a_1=1$, and the recurrence relation $a_n=20a_{n-1}+12a_{n-2}$ for all positive integers $n \geq 2$. If $\sqrt{\mathbb{S}}$ can be expressed in the form $\frac{a}{\sqrt{b}}$ where $a$ and $b$ are relatively prime positive integers. Determine the order pair $(a, b)$.

Thanks in advance.

REPLY [13 votes]: Hint: Consider the generating series $F(x)=\sum_{n=0}^{\infty}a_{n}x^{n}.$ Then, using the recurrence relation notice that \begin{align}
F(x)-20xF(x)-12x^{2}F(x) &= a_{0}+\left(a_{1}-20a_{0}\right)x+\sum_{n=2}^{\infty}\left(a_{n}-20a_{n-1}-12a_{n-2}\right)x^{n}  \\
&= 1-19x.
\end{align}
Hence for $x$ in the radius of convergence $$F(x)=\frac{1-19x}{1-20x-12x^{2}}.$$
(Remark:  The radius of convergence is $0.048$, or $\frac{1}{6}\left(2\sqrt{7}-5\right)$ which is close to $\frac{1}{19}$)
Now let $x=\frac{1}{10^2}$ and go from here.  
(You will indeed get a square as the numerator, the denominator is a prime number fairly close to $2000$.)