由 a(n+2) = 5/3a(n+1) - 2/3an可得,
a(n+2) - a(n+1) = 2/3[a(n+1)-an]
[a(n+2)-a(n+1)] / [a(n+1)-an] = 2/3
设 [a(n+1)-an] = bn
则 b(n+1)/bn = 2/3
b1 = a2 - a1 = 2/3
所以,bn = 2/3 *(2/3)^(n-1) = (2/3)^n
所以,a(n+1)-an = (2/3)^n
那么,an-a(n-1) = (2/3)^(n-1)
a(n-1)-a(n-2) = (2/3)^(n-2)
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a2 - a1 = 2/3
将以上n项相加,得,
a(n+1) - a1 = 2/3*[1-(2/3)^n]/(1-2/3)
a(n+1) - 1 = 2*[1-(2/3)^n]
a(n+1) = 3 - 2*(2/3)^n
即,an = 3 - 2*(2/3)^(n-1)