#30 Hua Luo Geng.

Algebra Level 3

{ a 1 = 5 a n + 1 = 1 + 3 a n 3 a n for n 2 \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 } \{a_n\} satisfies the recurrence relation above. Find a 100 a_{100} .

5 13 3 11 10 11 -\frac{13\sqrt3}{11}-\frac{10}{11} 1 5 -\frac{1}{5} 13 3 37 10 37 -\frac{13\sqrt3}{37}-\frac{10}{37}

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1 solution

Chew-Seong Cheong
Aug 22, 2017

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

a n + 1 = 1 + 3 a n 3 a n tan θ n + 1 = 1 + 3 tan θ n 3 tan θ n = 1 tan ( π 3 θ n ) = cot ( π 3 θ n ) = tan ( π 2 π 3 + θ n ) = tan ( π 6 + θ n ) \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}

θ n + 1 = π 6 + θ n θ n = ( n 1 ) π 6 + θ 1 where θ 1 = tan 1 5 a 100 = tan θ 100 = tan ( ( 100 1 ) π 6 + tan 1 5 ) = tan ( 16 1 2 π + tan 1 5 ) = tan ( π 2 + tan 1 5 ) Note that tan ( π x ) = tan x = tan ( π 2 tan 1 5 ) = cot ( tan 1 5 ) = 1 5 \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 ) f(x) , and found that f 6 ( x ) = x f^6(x)=x .

Kelvin Hong - 3 years, 9 months ago

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Done similar problem before, so I know the trick.

Chew-Seong Cheong - 3 years, 9 months ago

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Wow ,seems that you are familiar to many problems !

Kelvin Hong - 3 years, 9 months ago

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@Kelvin Hong I am on Brilliant every day. Problems solved 11,384. Current streak 1,183 Days. Solutions - 4074 total, ~26K upvotes. I think the answer given for #32 Hua Luo Geng is incorrect. I used an Excel spreadsheet and found that the answer is 1.43767E+90. Please check.

Chew-Seong Cheong - 3 years, 9 months ago

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@Chew-Seong Cheong Wow ,I hope I can also solve problems in Brilliant everyday ! Besides that, the question #32 hua luo geng can turns into binomial expansion . Or may you write your solution to me? Maybe I have something wrong in the question.

Kelvin Hong - 3 years, 9 months ago

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@Kelvin Hong In that case, your answer (-1) is correct. I am a moderator, I can check the answer actually. I edited this problem actually.

Chew-Seong Cheong - 3 years, 9 months ago

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@Chew-Seong Cheong Oh, have I given wrong information? I thought the expressions I have given isn't have any wrong .

Kelvin Hong - 3 years, 9 months ago

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@Kelvin Hong I have provided a solution for #32 and amended the problem wording

Chew-Seong Cheong - 3 years, 9 months ago

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@Chew-Seong Cheong Okay, thank you very much

Kelvin Hong - 3 years, 9 months ago

@Chew-Seong Cheong Otherwise, your solution is actually same as mine.

Kelvin Hong - 3 years, 9 months ago

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