设a1=1,a2=5/3,a(n+2)=(5/3)[a(n+1)-(2/3)an(n=1,2,3…）.(1)令bn=a(n+1)-an(n=1,2,…）,求数列{bn}的通项公式；（2）求数列{nan}的前n项和Sn

1.a(n+2)-a(n+1)=2/3 [a(n+1)-an]
[a(n+2)-a(n+1)]/[a(n+1)-an]=2/3
a(n+1)-an=(2/3)^n
bn=(2/3)^n
2.
a(n+1)-an=(2/3)^n
a2-a1=2/3
a3-a2=(2/3)^2
a4-a3=(2/3)^3
……
……
an-a(n-1)=(2/3)^(n-1)
叠加得
an-a1=2-(2/3)^(n-1)
an=3-(2/3)^(n-1)
nan=3n-n(2/3)^(n-1)
令Cn=n(2/3)^(n-1) 和为Tn
Tn=1*(2/3)^0+2*(2/3)^1+3*(2/3)^2+……+n*(2/3)^(n-1)
2/3Tn=1*(2/3)^1+2*(2/3)^2+3*(2/3)^3+……+(n-1)(2/3)^(n-1)+n*(2/3)^n
相减
1/3Tn=1*(2/3)^0+1*(2/3)^1+1*(2/3)^2+……+1*(2/3)^（n-1）-n*(2/3)^n
=3*[1-(2/3)^n]-n*(2/3)^n
Tn=9*[1-(2/3)^n]-3n*(2/3)^n
Sn=(1+3n)*n/2-Tn=(1+3n)*n/2-3*[1-(2/3)^n]+n*(2/3)^n