Question

、、数列{an}前n项算数平均数倒数1/(2n+1),求{an}通项公式

Answer 1

数列{an}前n项算数平均数为：2n+1

即数列{an}前n项的和为：n(2n+1)

即Sn=n(2n+1)

Sn=2n^2+n

S(n-1)=2(n-1)^2+n-1

=2n^2-4n+2+n-1

=2n^2-3n+1

an=Sn-S(n-1)

=2n^2+n-(2n^2-3n+1)

=2n^2+n-2n^2+3n-1

=4n-1