5
$\begingroup$

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.

  • 4
    @Qiaochu The value of $b$ suggests the contest may have been over for some time ;-).2011-07-25

1 Answers 1

13

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 $2000$.)