已知数列{an}的前n项和为Sn=n²+2n(1)求数列的通项公式an(2)Tn=1/a1a2+1/a2a3+1/a3a4+...+1/ana

问题描述:

已知数列{an}的前n项和为Sn=n²+2n(1)求数列的通项公式an(2)Tn=1/a1a2+1/a2a3+1/a3a4+...+1/ana

(1)Sn=n2+2nan=Sn-S(n-1)=n2+2n-[(n-1)2+2(n-1)]=2n+1(2)1/ana(n+1)=1/(2n+1)(2n+3)=[1/(2n+1)-1/(2n+3)]/2∴Tn=[1/3-1/5+1/5-1/7+1/7-1/9+……+1/(2n+1)-1/(2n+3)]/2=[1/3-1/(2n+3)]/2=n/[3(2n+3)]