已知数列{an}的前n项和为Sn,且有a1=2,3Sn=5an-an-1+3Sn-1。(1)求数列{an}的通项公式

2025-03-07 06:31:55
推荐回答(1个)
回答1:

1.
3Sn-3S(n-1)=5An-A(n-1)
3An=5An-A(n-1)
2An=A(n-1)
数列{An}是以2为首项,1/2为公比的等比数列
An=2×(1/2)^(n-1)=(1/2)^(n-2)

2.
Bn的前n项和Tn
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)