Question

已知数列前n项和Sn=2n^2-3n+1,n属于N,求它的通项公式？

Answer 1

当n=1时,A1=S1=2*1-3*1+1=0
当n大于等于2时,An=Sn-S（n-1）
=(2n^2-3n+1)-[2(n-1)^2-3(n-1)+1]
=(2n^2-3n+1)-(2n^2-7n+6)
=2n^2-3n+1-2n^2+7n-6
=4n-5
两种情况都要写~,5,4n-5,1,Sn-1=2(n-1)^2-3(n-1)+1
当n≥2时有：
an=Sn-Sn-1
=2n^2-3n+1-[2(n-1)^2-3(n-1)+1]　
=4n-5
当n=1时：S1=2-3+1=0
所以:
当n=1时，an=0
当n≥2时，an=4n-5,0,an=Sn-S(n-1)
=(2n^2-3n+1)-[2(n-1)^2-3(n-1)+1]
=(2n^2-3n+1)-(2n^2-7n+6)
=2n^2-3n+1-2n^2+7n-6
=4n-5,0,