Recursion Recursion Recursion

Algebra Level 5

x n + 1 = 2 x n + 3 y n y n + 1 = x n + 2 y n \begin{aligned} x_{n+1}&=&2x_n+3y_n \\ y_{n+1}&=&x_n+2y_n \end{aligned} When solving the above system of recursions with initial terms of x 0 = 2 , y 0 = 1 x_0 = 2, y_0 = 1 , the formula for x n x_n is equal to ( A B C ) n + 1 + ( A + B C ) n + 1 D \frac{(A-B\sqrt{C})^{n+1}+(A+B\sqrt{C})^{n+1}}{D} where A , A, B , B, C , C, and D D are positive integers with gcd ( A , B , D ) = 1 , \gcd(A,B,D)=1, and C C is square-free. Find the value of k = B A x k + k = D C y k \sum_{k=B}^Ax_k+\sum_{k=D}^Cy_k


The answer is 104.

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1 solution

x n + 2 = 2 x n + 1 + 3 y n + 1 = 2 ( 2 x n + 3 y n ) + 3 ( x n + 2 y n ) = 7 x n + 12 y n = 8 x n + 12 y n x n = 4 x n + 1 x n \begin{aligned} x_{n+2} & = 2 x_{n+1} + 3 y_{n+1} \\ & = 2 \left(2 x_{n} + 3 y_{n}\right) + 3 \left( x_{n} + 2 y_{n}\right) \\ & = 7 x_{n} + 12 y_{n} \\ & = 8 x_{n} + 12 y_{n} - x_n \\ & = 4x_{n+1} - x_n \end{aligned}

x n + 2 4 x n + 1 + x n = 0 \Rightarrow x_{n+2} - 4x_{n+1} + x_n = 0

Therefore, the characteristics equation is as follows:

x n + 2 4 x n + 1 + x n = 0 ( x 4 12 2 ) ( x 4 + 12 2 ) = 0 ( x 2 + 3 ) ( x 2 3 ) = 0 \begin{aligned} x_{n+2} - 4x_{n+1} + x_n & = 0 \\ \left(x - \frac{4-\sqrt{12}}{2} \right) \left(x - \frac{4 + \sqrt{12}}{2} \right) & = 0 \\ \left(x - 2 + \sqrt{3} \right) \left(x - 2 - \sqrt{3} \right) & = 0 \end{aligned}

x n = a ( 2 + 3 ) n + b ( 2 3 ) n x 0 = a + b = 2 x 1 = a ( 2 + 3 ) + b ( 2 3 ) = 2 ( a + b ) + 3 ( a b ) = 4 + 3 ( a b ) x 1 = 2 x 0 + 3 y 0 = 2 ( 2 ) + 3 ( 1 ) = 7 4 + 3 ( a b ) = 7 3 ( a b ) = 3 a b = 3 a + b = 2 a = 2 + 3 2 b = 2 a = 2 2 + 3 2 = 2 3 2 x n = a ( 2 + 3 ) n + b ( 2 3 ) n x n = ( 2 3 ) n + 1 + ( 2 + 3 ) n + 1 2 \begin{aligned} \Rightarrow x_n & = a(2 + \sqrt{3})^n + b(2 - \sqrt{3})^n \\ x_0 & = \color{#3D99F6} {a + b = 2} \\ x_1 & = a(2 + \sqrt{3}) + b(2 - \sqrt{3}) = 2(a+b) + \sqrt{3}(a-b) \\ & = 4 + \sqrt{3}(a-b) \\ x_1 & = 2x_0 + 3y_0 = 2(2) + 3(1) = 7 \\ \Rightarrow 4 + \sqrt{3}(a-b) & = 7 \\ \sqrt{3}(a-b) & = 3 \\ a - b & = \sqrt{3} \\ \color{#3D99F6} {a + b} & \color{#3D99F6} {= 2} \\ \Rightarrow a & = \frac{2+\sqrt{3}}{2} \\ \Rightarrow b & = 2 - a = 2 - \frac{2+\sqrt{3}}{2} = \frac{2-\sqrt{3}}{2} \\ x_n & = a(2 + \sqrt{3})^n + b(2 - \sqrt{3})^n \\ \Rightarrow x_n & = \frac{(2 - \sqrt{3})^{n+1} + (2 + \sqrt{3})^{n+1}}{2} \end{aligned}

A = 2 B = 1 C = 3 D = 2 \Rightarrow A = 2 \quad \Rightarrow B = 1 \quad \Rightarrow C = 3 \quad \Rightarrow D = 2

k = B A x k + k = D C y k = k = 1 2 x k + k = 2 3 y k = x 1 + x 2 + y 2 + y 3 = 7 + 26 + 15 + 56 = 104 \begin{aligned} \Rightarrow \sum_{k=B}^A x_k + \sum_{k=D}^C y_k & = \sum_{k=1}^2 x_k + \sum_{k=2}^3 y_k \\ & = x_1 + x_2 + y_2 + y_3 \\ & = 7 + 26 + 15 + 56 \\ & = \boxed{104} \end{aligned}

Moderator note:

Great! That's the standard approach of converting a double recurrence relation (involving only constants of a pervious term) into a single recurrence relation.

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