数列{an}的前n项和Sn=2n的平方-n+1,则an是多少_数学解答

数列{an}的前n项和Sn=2n的平方-n+1,则an是多少

人气：306 ℃　时间：2020-02-05 23:03:45

解答

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