设数列{an}的前n项和为Sn,且2an=Sn+2n+1(n∈N*).(Ⅰ)求a1,a2,a3;(Ⅱ)求证:数列{an+2}是等比数列;(Ⅲ)求数列{n•an}的前n项和Tn.
问题描述:
设数列{an}的前n项和为Sn,且2an=Sn+2n+1(n∈N*).
(Ⅰ)求a1,a2,a3;
(Ⅱ)求证:数列{an+2}是等比数列;
(Ⅲ)求数列{n•an}的前n项和Tn.
答
(本小题满分13分)(I) 由题意,当n=1时,得2a1=a1+3,解得a1=3.当n=2时,得2a2=(a1+a2)+5,解得a2=8.当n=3时,得2a3=(a1+a2+a3)+7,解得a3=18.所以a1=3,a2=8,a3=18为所求.…(3分)(Ⅱ)证明:因为2an...