数列的通项an和与前n项之和Sn的关系(1)已知数列{an}的前n项之和Sn=3n^2-2n,求an.问:此数列是什么数列?理由是为什么?(2)已知数列{an}的前n项之和Sn=2(n+1)^2,求an.问此数列为何数列,为什么?

Answer 1

此数列是等差数列sn=3n^2-2ns(n-1)=3(n-1)^2-2(n-1)=3n^2-6n+3-2n+2=3n^2-8n+5an=sn-s(n-1)=3n^2-2n-(3n^2-8n+5)=3n^2-2n-3n^2+8n-5=6n-5sn=2(n+1)^2s(n-1)=2n^2an=sn-s(n-1)=2(n+1)^2-2n^2=2n^2+4n+2-2n^2=4n+2此数...