数列{an}的前N项和为Sn,a1=1,an+1=2Sn(n∈N*). (Ⅰ)求数列{an}的通项an; (Ⅱ)求数列{nan}的前n项和Tn.
题目
数列{an}的前N项和为Sn,a1=1,an+1=2Sn(n∈N*).
(Ⅰ)求数列{an}的通项an;
(Ⅱ)求数列{nan}的前n项和Tn.
答案
(I)∵an+1=2Sn,∴Sn+1-Sn=2Sn,∴Sn+1Sn=3.又∵S1=a1=1,∴数列{Sn}是首项为1、公比为3的等比数列,Sn=3n-1(n∈N*).∴当n≥2时,an-2Sn-1=2•3n-2(n≥2),∴an=1,n=12•3n−2,n≥2(II)Tn=a1+2a2+3a3+...
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