设{an}为等比数例,Tn=na1+(n-1)a2…+2an-1+an,已知T1=1,T2=4, (1)求数列{an}的首项和公比; (2)求数列{Tn}的通项公式.

问题描述:

设{an}为等比数例,Tn=na1+(n-1)a2…+2an-1+an,已知T1=1,T2=4,
(1)求数列{an}的首项和公比;
(2)求数列{Tn}的通项公式.

(1)设等比数列{an}以比为q,则T1=a1,T2=2a1+a2=a1(2+q).∵T1=1,T2=4,∴a1=1,q=2.(2)设Sn=a1+a2+…+an.由(1)知an=2n-1.∴Sn=1+2+…+2n-1=2n-1∴Tn=na1+(n-1)a2+…+2an-1+an=a1+(a1+a2)+…+(a1+a...