Question

{an}的前n项和为Sn,a1=2,3Sn=5an-an-1（n-1都为下标）+3Sn-1（n-1都为下标） （n>=2

Answer 1

1）

3Sn=5an-a(n-1)+3S(n-1)

3[Sn-S(n-1)]=5an-a(n-1)

3an=5an-a(n-1)

2an=a(n-1)

an/a(n-1)=1/2

所以an是公比为1/2的等比数列

an=2*(1/2)^(n-1)=(1/2)^(n-2)

2)

bn=(2n-1)*(1/2)^(n-2)

Tn=1×2+3×1+5×(1/2)+……+(2n-1)×(1/2)^(n-2)

两边乘以2

2Tn=1×4+3×2+5×1+……+(2n-1)×(1/2)^(n-3)

错位相减

2Tn-Tn=4+2[2+1+1/2+……+(1/2)^(n-3)]-(2n-1)×(1/2)^(n-2)

=4+4×(1-(1/2)^(n-1))/(1-1/2)-(2n-1)/2^(n-2)

=12-(2n+3)/2^(n-2)