#30 Hua Luo Geng.

Algebra Level 3

$\large \begin{cases} a_1 =5 \\ a_{n+1}=\dfrac{1+\sqrt3 a_n}{\sqrt3 -a_n} & \text{for } n \ge 2 \end{cases}$

A sequence $\{a_n\}$ satisfies the recurrence relation above. Find $a_{100}$ .

-\frac{13\sqrt3}{11}-\frac{10}{11}

-\frac{1}{5}

-\frac{13\sqrt3}{37}-\frac{10}{37}

1 solution

Chew-Seong Cheong
Aug 22, 2017

Let $a_n = \tan \theta_n$ , then we have, for $n \ge 2$ ,

$\begin{aligned} a_{n+1} & = \frac {1+\sqrt 3 a_n}{\sqrt 3 - a_n} \\ \tan \theta_{n+1} & = \frac {1+\sqrt 3 \tan \theta_n}{\sqrt 3 - \tan \theta_n} = \frac 1{\tan \left(\frac \pi 3 - \theta_n\right)} = \cot \left(\frac \pi 3 - \theta_n\right) = \tan \left(\frac \pi 2 - \frac \pi 3 + \theta_n \right) = \tan \left(\frac \pi 6 + \theta_n \right) \end{aligned}$

$\begin{aligned} \implies \theta_{n+1} & = \frac \pi 6 + \theta_n \\ \theta_n & = \frac {(n-1)\pi} 6 + \color{#3D99F6}\theta_1 & \small \color{#3D99F6} \text{where }\theta_1 = \tan^{-1} 5 \\ \implies a_{100} & = \tan \theta_{100} \\ & = \tan \left(\frac {(100-1)\pi} 6 + \tan^{-1} 5\right) \\ & = \tan \left(16\frac 12 \pi + \tan^{-1} 5\right) \\ & = \tan \left(\frac \pi 2 + \tan^{-1} 5\right) & \small \color{#3D99F6} \text{Note that }\tan (\pi -x) = - \tan x \\ & = - \tan \left(\frac \pi 2 - \tan^{-1} 5\right) \\ & = - \cot \left(\tan^{-1} 5\right) \\ & = \boxed{- \dfrac 15} \end{aligned}$

Awesome solution, I don't even think of that! I use $f(x)$ , and found that $f^6(x)=x$ .