已知Sn为数列{an}的前项n和,且Sn=2an+n^2-3n-2(n∈N*),
求证：数列{an－2n}是等比数列,并求数列{an}的 通项公式
【解】Sn=2an＋n2－3n-2,则S1=2a1＋1－3-2,a1=4.
Sn=2an+n^2-3n S(n-1)=2a(n-1)+(n-1)^2-3(n-1)
Sn-S(n-1)=2an-2a(n-1)+n^2-(n-1)^2-3n+3(n-1)
即an =2an-2a(n-1)+2n-4
an=2a(n-1)-2n+4
an-2n=2a(n-1)-4n+4=2a(n-1)-4(n-1)
an-2n =2[a(n-1)-2(n-1)]
(an-2n)/[a(n-1)-2(n-1)]=2
数列{an－2n}是等比数列,a1-2=2,
an－2n=2•2^(n-1)
an=2^n+2n.