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.

  • 0
    Is this homework? What have you tried?2011-07-25
  • 10
    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.2011-07-25
  • 2
    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?2011-07-25
  • 0
    @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...2011-07-25
  • 4
    @Qiaochu The value of $b$ suggests the contest may have been over for some time ;-).2011-07-25

1 Answers 1