Question

数列{an}的前n项和记为Sn,a1=t,点(Sn,an+1)在直线y=3x+1上,n∈N*.

(1)当实数t为何值时,数列{an}是等比数列?

(2)在(1)的结论下,设bn=log4an+1,cn=an+bn,Tn是数列{cn}的前n项和,求Tn.

 

Answer 1

【答案】

(1) t=1 (2) Tn=+

【解析】

解:(1)∵点(Sn,an+1)在直线y=3x+1上,

∴an+1=3Sn+1,an=3Sn-1+1(n>1),

an+1-an=3(Sn-Sn-1)=3an,

∴an+1=4an,

a2=3S1+1=3a1+1=3t+1,

∴当t=1时,a2=4a1,数列{an}是等比数列.

(2)在(1)的结论下,an+1=4an,an+1=4n,

bn=log4an+1=n,

cn=an+bn=4n-1+n,

Tn=c1+c2+…+cn

=(40+1)+(41+2)+…+(4n-1+n)

=(1+4+42+…+4n-1)+(1+2+3+…+n)

=+.