x n + 1 y n + 1 = = 2 x n + 3 y n x n + 2 y n When solving the above system of recursions with initial terms of x 0 = 2 , y 0 = 1 , the formula for x n is equal to D ( A − B C ) n + 1 + ( A + B C ) n + 1 where A , B , C , and D are positive integers with g cd ( A , B , D ) = 1 , and C is square-free. Find the value of k = B ∑ A x k + k = D ∑ C y k
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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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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 + 1 2 y n = 8 x n + 1 2 y n − x n = 4 x n + 1 − x n
⇒ x n + 2 − 4 x n + 1 + x n = 0
Therefore, the characteristics equation is as follows:
x n + 2 − 4 x n + 1 + x n ( x − 2 4 − 1 2 ) ( x − 2 4 + 1 2 ) ( x − 2 + 3 ) ( x − 2 − 3 ) = 0 = 0 = 0
⇒ x n x 0 x 1 x 1 ⇒ 4 + 3 ( a − b ) 3 ( a − b ) a − b a + b ⇒ a ⇒ b x n ⇒ x n = a ( 2 + 3 ) n + b ( 2 − 3 ) n = a + b = 2 = a ( 2 + 3 ) + b ( 2 − 3 ) = 2 ( a + b ) + 3 ( a − b ) = 4 + 3 ( a − b ) = 2 x 0 + 3 y 0 = 2 ( 2 ) + 3 ( 1 ) = 7 = 7 = 3 = 3 = 2 = 2 2 + 3 = 2 − a = 2 − 2 2 + 3 = 2 2 − 3 = a ( 2 + 3 ) n + b ( 2 − 3 ) n = 2 ( 2 − 3 ) n + 1 + ( 2 + 3 ) n + 1
⇒ A = 2 ⇒ B = 1 ⇒ C = 3 ⇒ 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 + 2 6 + 1 5 + 5 6 = 1 0 4