A Dubicious Recursion?
x 1 x 2 x 3 x 4 x n = 2 1 1 , = 3 7 5 , = 4 2 0 , = 5 2 3 , and = x n − 1 − x n − 2 + x n − 3 − x n − 4 when n ≥ 5 ,
Given the recurrence relation above, find the value of x 5 3 1 + x 7 5 3 + x 9 7 5 .
This an old Olympiad problem.
The answer is 898.
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2 solutions
We are given that x n = x n − 1 − x n − 2 + x n − 3 − x n − 4 .
This means that x n + 1 = x n − x n − 1 + x n − 2 + x n − 3 = ( x n − 1 − x n − 2 + x n − 3 − x n − 4 ) − x n − 1 + x n − 2 + x n − 3 = − x n − 4
So in general x n = − x n − 5 . Using the given relation, x 5 = 2 6 7 so the sequence 211, 375, 420, 523, 267, -211, -375, -420, -523, -267 repeats. All terms corresponding to values of n that have the same remainder when divided by 10 have the same value in the sequence.
Therefore x 5 3 1 = x 1 = 2 1 1
x 7 5 3 = x 3 = 4 2 0
x 9 7 5 = x 5 = 2 6 7
So x 5 3 1 + x 7 5 3 + x 9 7 5 = 2 1 1 + 4 2 0 + 2 6 7 = 8 9 8
x n + 4 = x n + 3 − x n + 2 + x n + 1 − x n
x n + 5 = x n + 4 − x n + 3 + x n + 2 − x n + 1
Adding the above 2 equations, x n + 4 + x n + 5 = x n + 4 − x n .
⇒ x n + 5 = − x n ⇒ x n + 1 0 = − x n + 5 = x n .
Hence, the sequence has a period of 10. Therefore,
x 5 3 1 + x 7 5 3 + x 9 7 5
= x 1 + x 3 + x 5
= x 1 + x 3 + ( x 4 − x 3 + x 2 − x 1 )
= x 2 + x 4 = 3 7 5 + 5 2 3 = 8 9 8 .