Sn=n-2an ,Sn-1=(n-1)-2an-1(n大于1)
做差an=1-2an-2an-1
3an-3=2an-1-2
(an-1)/[a(n-1)-1]=2/3是常数,经检验,a1=1/3,a2=5/9,满足上式,所以{an-1}是等比数列
an-1=(-2/3)*(2/3)^n-1=-(2/3)^n
an=1-(2/3)^n
bn=n(1+an)=[2-(2/3)^n]*n=2n-n(2/3)^n
由错项相减法有Tn=n(n+1)+(n+3)*(2/3)^n-3
Sn=n-2an ,Sn-1=(n-1)-2an-1(n大于1)
做差an=1-2an-2an-1
3an-3=2an-1-2
(an-1)/[a(n-1)-1]=2/3是常数,经检验,a1=1/3,a2=5/9,满足上式,所以{an-1}是等比数列
an-1=(-2/3)*(2/3)^n-1=-(2/3)^n
an=1-(2/3)^n
bn=n(1+an)=[2-(2/3)^n]*n=2n-n(2/3)^n
由错项相减法有Tn=n(n+1)+(n+3)*(2/3)^n-3