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Algebra Level 4

A certain sequence is such that its first term x 1 = 2017 x_1 = 2017 and for n ≥ 1 n \ge 1 , x n + 1 = ( 2 + 1 ) x n − 1 2 + 1 + x n x_{n+1} = \dfrac{\left( \sqrt{2} + 1 \right)x_{n} - 1}{\sqrt{2} + 1 + x_n }

If x 2015 = − a b x_{2015} = -\dfrac{a}{b} , where a a and b b are coprime positive integers, what is a + 2 b a+2b ?

For more problems like this, try this set .


The answer is 3025.

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Christian Daang by Christian Daang , 20, Philippines

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2 solutions

Chew-Seong Cheong
Jul 24, 2017

x n + 1 = ( 2 + 1 ) x n − 1 2 + 1 + x n Let x n = tan ⁡ θ n tan ⁡ θ n + 1 = ( 2 + 1 ) tan ⁡ θ n − 1 2 + 1 + tan ⁡ θ n Note: tan ⁡ 3 π 8 = 2 + 1 = tan ⁡ 3 π 8 tan ⁡ θ n − 1 tan ⁡ 3 π 8 + tan ⁡ θ n = − 1 tan ⁡ ( θ n + 3 π 8 ) Note: tan ⁡ ( π − x ) = − tan ⁡ x = 1 tan ⁡ ( 5 π 8 − θ n ) = cot ⁡ ( 5 π 8 − θ n ) Note: cot ⁡ ( π 2 − x ) = tan ⁡ x = tan ⁡ ( θ n − π 8 ) ⟹ θ n + 1 = θ n − π 8 An arithmetic progression θ n = θ 1 − ( n − 1 ) π 8 where θ 1 = tan ⁡ − 1 2017 ⟹ x n = tan ⁡ ( θ 1 − ( n − 1 ) π 8 ) x 2015 = tan ⁡ ( θ 1 − ( 2014 ) π 8 ) = tan ⁡ ( θ 1 − 3 π 4 ) = tan ⁡ ( θ 1 + π 4 ) = tan ⁡ θ 1 + 1 1 − tan ⁡ θ 1 = − 2018 2016 = − 1009 1008 \begin{aligned} \color{#3D99F6}x_{n+1} & = \frac {(\sqrt 2+1){\color{#3D99F6}x_n} - 1}{\sqrt 2+1+{\color{#3D99F6}x_n}} & \small \color{#3D99F6} \text{Let }x_n = \tan \theta_n \\ \color{#3D99F6}\tan \theta_{n+1} & = \frac {({\color{#D61F06}\sqrt 2+1}){\color{#3D99F6}\tan \theta_n} - 1}{{\color{#D61F06}\sqrt 2+1}+{\color{#3D99F6}\tan \theta_n}} & \small \color{#D61F06} \text{Note: } \tan \frac {3\pi}8 = \sqrt 2+1 \\ & = \frac {{\color{#D61F06}\tan \frac {3\pi}8}\tan \theta_n - 1}{{\color{#D61F06}\tan \frac {3\pi}8}+\tan \theta_n} \\ & = - \frac 1{\color{#3D99F6}\tan \left(\theta_n + \frac {3\pi}8 \right)} & \small \color{#3D99F6} \text{Note: } \tan (\pi -x) = - \tan x \\ & = \frac 1{\color{#3D99F6}\tan \left(\frac {5\pi}8 - \theta_n\right)} \\ & =\color{#3D99F6} \cot \left(\frac {5\pi}8 - \theta_n\right) & \small \color{#3D99F6} \text{Note: } \cot \left(\frac \pi 2 - x\right) = \tan x \\ & = \color{#3D99F6} \tan \left( \theta_n - \frac \pi 8\right) \\ \implies \theta_{n+1} & = \theta_n - \frac \pi 8 & \small \color{#3D99F6} \text{An arithmetic progression} \\ \theta_n & = \theta_1 - (n-1)\frac \pi 8 & \small \color{#3D99F6} \text{where }\theta_1 = \tan^{-1} 2017 \\ \implies x_n & = \tan \left(\theta_1 - (n-1)\frac \pi 8 \right) \\ x_{2015} & = \tan \left(\theta_1 - (2014)\frac \pi 8 \right) \\ & = \tan \left(\theta_1 - \frac {3\pi}4 \right) \\ & = \tan \left(\theta_1 + \frac \pi 4 \right) \\ & = \frac {\tan \theta_1+1}{1-\tan \theta_1} \\ & = - \frac {2018}{2016} = - \frac {1009}{1008} \end{aligned}

⟹ a + 2 b = 1009 + 2 ( 1008 ) = 3025 \implies a + 2b = 1009 + 2(1008) = \boxed{3025}

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The Solution is really great. Fantastic. ^^

Christian Daang - 3 years, 10 months ago

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Glad that you like it.

Chew-Seong Cheong - 3 years, 10 months ago

Sir, i am not getting how did you come up with fifth last step

Shefali Sharma - 2 years, 8 months ago

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Note that for angle 2 n π + θ ≡ θ 2n\pi + \theta \equiv \theta , where n n is an integer. 2014 8 π ≡ 1007 4 π ≡ 250 π + 7 4 π ≡ 7 4 π \frac {2014}8 \pi \equiv \frac {1007}4 \pi \equiv 250\pi + \frac 74\pi \equiv \frac 74 \pi . And we note that tan ⁡ ( π + θ ) = tan ⁡ θ \tan (\pi + \theta) = \tan \theta . Therefore, tan ⁡ 7 4 π = tan ⁡ ( π + 3 4 π ) = tan ⁡ 3 4 π \tan \frac 74 \pi = \tan \left(\pi + \frac 34\pi\right) = \tan \frac 34 \pi .

Chew-Seong Cheong - 2 years, 8 months ago

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Oh I got it, Thank you so much for replying

Shefali Sharma - 2 years, 8 months ago
Christian Daang
Jul 23, 2017

Using the iterative formula above,

x n + 2 = ( 2 + 1 ) ( x n + 1 ) − 1 ( 2 + 1 ) + x n + 1 = ( 2 + 1 ) ( ( 2 + 1 ) ( x n ) − 1 ( 2 + 1 ) + x n ) − 1 ( 2 + 1 ) + ( 2 + 1 ) ( x n ) − 1 ( 2 + 1 ) + x n ⋅ 2 + 1 + x n 2 + 1 + x n = ( 3 + 2 2 ) ( x n ) − ( 1 + 2 ) − ( 2 + 1 + x n ) ( 3 + 2 2 ) + ( 1 + 2 ) ( x n ) + ( 2 + 1 ) ( x n ) − 1 = ( 2 + 2 2 ) ( x n ) − ( 2 + 2 2 ) ( 2 + 2 2 ) ( x n ) + ( 2 + 2 2 ) = x n − 1 x n + 1 \begin{aligned} \displaystyle x_{n+2} &= \dfrac{\left( \sqrt{2} + 1 \right)\left( \color{#D61F06}{x_{n+1}} \right) - 1}{\left( \sqrt{2} + 1 \right) +\color{#D61F06}{x_{n+1}} } \\ &= \dfrac{\left( \sqrt{2} + 1 \right)\left( \color{#D61F06}{\dfrac{\left( \sqrt{2} + 1 \right)(x_{n}) - 1}{\left( \sqrt{2} + 1 \right) + x_n }} \right) - 1}{\left( \sqrt{2} + 1 \right) + \color{#D61F06}{\dfrac{\left( \sqrt{2} + 1 \right)(x_{n}) - 1}{\left( \sqrt{2} + 1 \right) + x_n }} } \cdot \dfrac{ \sqrt{2} + 1 + x_n }{ \sqrt{2} + 1 + x_n } = \dfrac{\left(3 + 2\sqrt{2}\right)(x_n) - \left(1 + \sqrt{2}\right) - \left( \sqrt{2} + 1 + x_n \right)}{\left(3 + 2\sqrt{2}\right) + \left(1 + \sqrt{2}\right)(x_n) + \left( \sqrt{2} + 1\right)(x_n) - 1} \\ &= \dfrac{ \left( 2 + 2\sqrt{2} \right)(x_n) - \left(2 + 2\sqrt{2} \right)}{ \left( 2 + 2\sqrt{2} \right)(x_n) + \left(2 + 2\sqrt{2} \right)} \\ &= \dfrac{x_n - 1}{x_n + 1} \end{aligned}

Now, x n + 4 = ( x n + 2 ) − 1 ( x n + 2 ) + 1 = ( x n − 1 x n + 1 ) − 1 ( x n − 1 x n + 1 ) + 1 = − 1 x n x_{n+4} = \dfrac{ \left( \color{#E81990}{x_{n+2} } \right) - 1 }{ \left( \color{#E81990}{ x_{n+2} } \right) + 1 } = \dfrac{ \left( \color{#E81990}{ \dfrac{x_n - 1}{x_n + 1} } \right) - 1 }{ \left( \color{#E81990}{ \dfrac{x_n - 1}{x_n + 1} }\right) + 1 } = -\dfrac{1}{x_n}

∴ x n + 8 = − 1 x n + 4 = − 1 ( − 1 x n ) = x n \therefore x_{n+8} = -\dfrac{1}{ \color{#20A900}{x_{n+4}}} = -\dfrac{1}{ \left( \color{#20A900}{ -\dfrac{1}{x_n} }\right) } = x_n

⟹ x 2015 = x 2007 = x 1999 = ⋯ = x 7 = − 1 x 3 = − 1 ( x 1 − 1 x 1 + 1 ) = − 2018 2016 = − 1009 1008 ⟹ a = 1009 , b = 1008 ∴ a + 2 b = 3025 \begin{aligned} \implies x_{2015} = x_{2007} = x_{1999} = \cdots = x_7 &= -\dfrac{1}{ \color{cyan}{x_3}} \\ &= -\dfrac{1}{ \left( \color{cyan}{ \dfrac{x_1 - 1}{x_1 + 1}} \right) } \\ &= -\dfrac{2018}{2016} \\ &= -\dfrac{1009}{1008} \implies a = 1009 \ , \ b = 1008 \\ \therefore \boxed{a + 2b = 3025} \end{aligned}

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