数列{an}的前n项和Sn=2n的平方-n+1,则an是多少
问题描述：
数列{an}的前n项和Sn=2n的平方-n+1,则an是多少
答
Sn=2n^2-n+1
Sn-1=2(n-1)^2-(n-1)+1
=2(n^2-2n+1)-n+1+1
=2n^2-4n+2-n+2
=2n^2-5n+4
Sn-Sn-1=An= 2n^2-n+1-(2n^2-5n+4)
= 2n^2-n+1-2n^2+5n-4
=4n-3