问题描述：
设a1=1,a2=5/3,a(n+2)=5/3 a(n+1)- 2/3 an(n属于N*) ,令bn=a(n+1) -an,求数列{bn}的通项公式
(2)求数列{n*an}的前n项和Sn.
问题解答：
我来补答
a(n+2)=(5/3)a(n+1)-(2/3)an
a(n+2)-a(n+1)=(2/3)[a(n+1)-an]
bn=a(n+1)-an
b(n+1)=(2/3)bn
bn=b1*(2/3)^(n-1)
bn=(2/3)^n
a(n+1)-an=(2/3)^n
a2-a1=(2/3)^1
a3-a2=(2/3)^2
a4-a3=(2/3)^3
....
a(n+1)-an=(2/3)^n
全部相加
a(n+1)-a1=2[1-(2/3)^n]
a(n+1)=2[1-(2/3)^n]+1
a(n+1)=3-2(2/3)^n
an=3-2(2/3)^(n-1)
nan=3n-2n(2/3)^(n-1)
Sn=3(1+n)n/2-2[1*(2/3)^0+2*(2/3)^1+3*(2/3)^2+……+n(2/3)^(n-1)]
令S1=1*(2/3)^0+2*(2/3)^1+3*(2/3)^2+……+n(2/3)^(n-1)
(2/3)S1=1*(2/3)^1+2*(2/3)^2+……+n(2/3)^n
两式相减
S1/3=(2/3)^0+(2/3)^1+(2/3)^2+……+(2/3)^(n-1)-n(2/3)^n
S1/3=3[1-(2/3)^n]-n(2/3)^n
S1=9[1-(2/3)^n]-3n(2/3)^n
Sn=3(1+n)n/2-18[1-(2/3)^n]+6n(2/3)^n
