己知数列{an}的前n项和Sn=3n^2-2n,求证:数列{an}成等差数列,并求其首项、公差、通项公式.

an=sn-s(n-1)
=3n²-2n-3(n-1)²+2(n-1)
=3n²-2n-3n²+6n-3+2n-2
=6n-5
同理a(n+1)=s(n+1)-sn=6n+1
a(n+1)-an=6
所以{an}为等差数列,公差为6
通项公式为an=6n-5
首项a1=6-5=1