Question

1. 设数列{an}的前n项和为Sn=2n^2,{bn}为等比数列,且a1=b1,b2(a2--a1)=b1,设cn=a

Answer 1

Tn=-11/9+(2/3×n-1/9)×4^n

Tn=（2-1）×4^（1-1)+（2×2-1）×4^(2-1)……(2n-1) ×4^(n-1)

为①式

4Tn=（2-1）×4^(2-1）+ （2×2-1）×4^(2-1）……(2n-1) ×4^n

为②式

则①-②得

-3Tn=1-2×4^1-2×4^2-……-2×4^(n-1)-(2n-1) ×4^n

-3Tn=1-2[(4-4^n)/(1-4)]-(2n-1) ×4^n

-3Tn=11/3-2/3×4^n-(2n-1) ×4^n

Tn=-11/9+(2/3×n-1/9)×4^n