设数列〔an〕满足a1=1,a2=5/3(3分之5）,an+2=5/3an+1－2/3an,(n属于N※）

(1)
a(n+2)=(5/3)a(n+1)-(2/3)an
a(n+2)-a(n+1) = (2/3)(a(n+1) -an)
{a(n+1) -an} 是等比数列, q=2/3
a(n+1) -an = (2/3)^(n-1).(a2-a1)
= (2/3)^n
bn = a(n+1) -an = (2/3)^n

a(n+1) - an = (2/3)^n
an -a(n-1) = (2/3)^(n-1)
an - a1 = (2/3)+(2/3)^2+.+(2/3)^(n-1)
= (2/3)(1- (2/3)^(n-1) )/(1-2/3)
= 2(1- (2/3)^(n-1) )
an = 3 - 2(2/3)^(n-1)

(2)
nan = 3n - 2[n(2/3)^(n-1)]
let
S = 1.(2/3)^0+2.(2/3)^1+...+n(2/3)^(n-1) (1)
(2/3)S = 1.(2/3)^1+2.(2/3)^2+...+n(2/3)^n (2)
(1)-(2)
(1/3)S = [1+2/3+...+(2/3)^(n-1)]- n(2/3)^n
= 3[1- (2/3)^n] - n(2/3)^n
S =9[1- (2/3)^n] - 3n(2/3)^n
= 9 - (3n-9)(2/3)^n
Sn = 3n(n+1)/2 - 2S
=n(n+1) - 18 + 2(3n-9)(2/3)^n