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.

  • 1
    $\begingroup$ Is this homework? What have you tried? $\endgroup$ – JavaMan Jul 25 '11 at 17:43
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    $\begingroup$ Just my opinion, but I think that this would read a bit better without the first two sentences (Am I really respected? Maybe, maybe not, but it's hardly relevant. Are two 'pleases' better than one? Why thank in advance?) Also, having an actual question somewhere in the body would help. Apologies for the nitpicking. $\endgroup$ – George Lowther Jul 25 '11 at 17:53
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    $\begingroup$ Also, what is the motivation? Why do you think that some fairly arbitrary looking infinite sequence should converge to something of the form $a^2/b$? Have you tried just solving the recurrence relation and summing the resulting geometric series? $\endgroup$ – George Lowther Jul 25 '11 at 17:56
  • $\begingroup$ @George: it looks like a contest problem to me. They are often phrased in such ways. This had better not be from an ongoing contest... $\endgroup$ – Qiaochu Yuan Jul 25 '11 at 18:12
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    $\begingroup$ @Qiaochu The value of $b$ suggests the contest may have been over for some time ;-). $\endgroup$ – whuber Jul 25 '11 at 22:07

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 $2000$.)


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