遞推數列求通項

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先求不動點,令 x = (2x-4)/(x+6)

==> x^2 + 4x + 4 = 0
==>(x + 2)^2 = 0
==> x = - 2

因有唯一不動點 - 2 , 故{ 1 / (an + 2) } 是等差數列。

即 :

a n+1 + 2 = (2an - 4) / (an + 6) + 2 = (4an + 8) / (an + 6)

1 / (a n+1 + 2) = (an + 6) / (4an + 8) = 1 / (an + 2) + 1/4

公差 = 1/4

故 1 / (an + 2) = 1 / (a1 + 2) + (n - 1)(1/4)

1 / (an + 2) = 1 / (2+2) +(n - 1)(1/4)

1 / (an + 2) = n/4

an + 2 = 4/n

an = 4/n - 2

an = 2(2 - n)/n

題目錯了
"an=2"這句應該是"a1/a0(?) = 2"