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斐波那契数列通项公式的几种求法
人气:453 ℃ 时间:2020-02-06 03:22:40
解答
 1.x(1) = 1, x(2) = 1, x(3) = x(1) + x(2) = 2, ..., x(n) = x(n-1) + x(n-2), ...这就是斐波那契数列设x(n) + a1 x(n-1) = a2(x(n-1) + a1 x(n-2))a2 - a1 = 1a2 X a1 = 1{x(n) + a1 x(n-1)} 就是等比...
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