已知数列{an}的前n项和公式为sn=2n^2-3n-1,则通项公式an=?
问题描述：
已知数列{an}的前n项和公式为sn=2n^2-3n-1,则通项公式an=?
答
s(n+1)=2(n+1)^2-3(n+1)-1,
a(n+1)=s(n+1)-s(n)=2(n+1)^2-3(n+1)-2n^2+3n
=2(2n+1)-3=4n-1=4(n+1)-5,
a(1)=s(1)=2-3-1=-2,
a(n)=4n-5,n=2,3,...
答
sn=2n^2-3n-1
s(n+1)=2(n+1)^2-3(n+1)-1
=2(n^2+1+2n)-3n-4
=2n^2+2+4n-3n-4
=2n^2+n-2
s(n-1)=2(n-1)^2-3(n-1)-1
=2(n^2+1-2n)-3n+2
=2n^2+2-4n-3n+2
=2n^2-7n+4
an=sn-s(n-1)=4n-5
a(n+1)=s(n+1)-sn=4n-1=4(n+1)-5
所以通向公式an=4n-5