The
n
th Lucas number is defined such that
L
n
=
L
n
−
1
+
L
n
−
2
, where
L
1
=
1
and
L
2
=
3
.
Similarly, the
n
th Fibonacci number is defined such that
F
n
=
F
n
−
1
+
F
n
−
2
, where
F
1
=
1
and
F
2
=
1
.
Let
Q
n
be the ratio
F
n
L
n
.
With simple calculation we find that
Q
1
=
1
1
=
1
,
Q
2
=
1
3
=
3
,
.
.
.
Find n → ∞ lim Q n .
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用特征根法 由于两个数列的递推式相同,故其特征根相同。 由 f ( n ) = f ( n − 1 ) + f ( n − 2 ) 可得特征方程 x 2 = x + 1 ,解得 x 1 = 2 1 − 5 , x 2 = 2 1 + 5 。 可设 L ( n ) = a ⋅ x 1 n + b ⋅ x 2 n F ( n ) = c ⋅ x 1 n + d ⋅ x 2 n
将 L ( 1 ) = 1 , L ( 2 ) = 3 , F ( 1 ) = 1 , F ( 2 ) = 1 代入,解得
c = − 5 1 , d = 5 1 a = 1 , b = 1
令 X = x 1 n , Y = x 2 n , t = Y / X 则
Q n = c ⋅ X + d ⋅ Y a ⋅ X + b ⋅ Y = c + d ⋅ Y / X a + b ⋅ Y / X = c + d t a + b t 当 n → ∞ 时, t → ∞ 可用 L ′ H o s p i t a l ′ s r u l e ,则 lim n → + ∞ Q n = lim t → + ∞ c + d t a + b t = d b = 5 。
thus the answer is 5
Python:
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what a method of brutal force!
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quite simple if u know
Ln = F(n-1)+F(n+1)
hence
Qn= F(n-1)+F(n+1) /Fn
Qn=2F(n-1) /Fn + 1
Qn=2(1/golden ration) + 1 .at inf
hence ANS = sqrt5