The 3003rd element

Algebra Level 4

Let x 0 , x 1 , x 2 , x_0, x_1, x_2,\ldots be a sequence of real numbers satisfying the recursion,

x n = x n 1 + x n + 1 x_n = x_{n-1} + x_{n+1}

for n = 1 , 2 , 3 , n = 1,2,3,\ldots .

If x 0 = 23 x_0 = 23 , what is x 3003 x_{3003} ?


The answer is -23.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Geoff Pilling
Aug 23, 2016

If we let x 1 = y x_1 = y , then

  • x 2 = y 23 x_2 = y-23
  • x 3 = 23 x_3 = -23
  • x 4 = y x_4 = -y
  • x 5 = 23 y x_5 = 23-y
  • x 6 = 23 x_6 = 23
  • x 7 = y x_7 = y

And the sequence repeats with a period of 6 6 .

So, in general, for n n divisible by 3 3 , x n = ( 1 ) n / 3 × 23 x_n = (-1)^{n/3} \times 23

So, x 3003 = 23 x_{3003} = \boxed{-23}

Chew-Seong Cheong
Aug 23, 2016

From the linear recurrent relation x n = x n 1 + x n + 1 x_n = x_{n-1} + x_{n+1} , the characteristic equation is as follows:

r 2 r + 1 = 0 r = 1 ± 3 i 2 = e ± π 3 i the 6th root of unity \begin{aligned} r^2 - r + 1 & = 0 \\ \implies r & = \frac {1 \pm \sqrt{3}i}2 \\ & = \color{#3D99F6}{e^{\pm \frac \pi 3 i}} & \small \color{#3D99F6}{\text{the 6th root of unity}} \end{aligned}

a n = a 1 e n π 3 i + a 2 e n π 3 i a 0 = a 1 + a 2 \begin{aligned} \implies a_n & = a_1 e^{\frac {n\pi}3i} + a_2 e^{- \frac {n\pi}3i} \\ a_0 & = a_1 + a_2 \end{aligned}

a 1 + a 2 = 23 \begin{aligned} \implies \color{#3D99F6}{a_1 + a_2} & \color{#3D99F6}{ = 23} \end{aligned}

a 3003 = a 1 e 3003 π 3 i + a 2 e 3003 π 3 i = a 1 e 1001 π i + a 2 e 1001 π i = a 1 e π i + a 2 e π i = a 1 a 2 = 23 \begin{aligned} \implies a_{3003} & = a_1 e^{\frac {3003 \pi}3i} + a_2 e^{- \frac {3003 \pi}3i} \\ & = a_1 e^{1001 \pi i} + a_2 e^{- 1001 \pi i} \\ & = a_1 e^{\pi i} + a_2 e^{- \pi i} \\ & = \color{#3D99F6}{ -a_1 - a_2} \\ & = \boxed{\color{#3D99F6}{-23}} \end{aligned}

Nice approach! :)

Geoff Pilling - 4 years, 9 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...