Question

已知数列{an}的前n项和为Sn=n^2+n,①求数列{an}的通项公式 ②若bn=(1/2)^an,求数列{bn}的前n项和Tn 求详细

求 Tn的详细过程

Answer 1

解：由已知条件可知
an=Sn-S（n-1）=n^2+n-(n-1)^2-(n-1)=2n
bn=(1/2)^an=(1/2)^(2n)=(1/4)^n
由数列｛bn｝通项可知，bn是以1/4为首项，公比为1/4的等比数列
则Tn={1/4[1-(1/4)^n]}/(1-1/4)=[1-(1/4)^n]/3

Answer 2

解：(1)、∵Sn=n²+n
∴S(n-1)=(n-1)²+(n-1)=n²-n
∴an=Sn-S(n-1)
=(n²+n)-(n²-n)
=2n
(2)、∵bn=(1/2)^an
∴bn=(1/2)^2n
∴Tn=(1/2)^2+(1/2)^4+……+(1/2)^2n
∴4Tn=1+(1/2)^2+……+(1/2)^(2n-2)
∴3Tn=4Tn-Tn=[(1/2)^2+(1/2)^4+……+(1/2)^2n]-[1+(1/2)^2+(1/2)^4+……+(1/2)^2n]
=1-(1/2)^2n
∴Tn=[1-(1/2)^2n]/3

Answer 3

Sn-Sn-1=an=2n
bn=(1/2)^an=(1/2)^2n=1/4^n
等比数列
Tn=1/4*(1-1/4^n)/(1-1/4)=1/3*(1-1/4^n)

Answer 4