已知数列{an}的前项和是Sn=n平方+n+1,则数列的通项an等于几?
问题描述:
已知数列{an}的前项和是Sn=n平方+n+1,则数列的通项an等于几?
答
an=sn-s(n-1)
=n^2+n+1-(n-1)^2-(n-1)-1
=n^2+n+1-n^2+2n-1-n+1-1
=2n