已知数列{an}的前n项和,Sn=n2+2n+1. (1)求数列{an}的通项公式an; (2)记Tn=1/a1a2+1/a2a3+…+1/anan+1,求Tn.

问题描述:

已知数列{an}的前n项和,Sn=n2+2n+1
(1)求数列{an}的通项公式an
(2)记Tn=

1
a1a2
+
1
a2a3
+…+
1
anan+1
,求Tn

(I)当n=1时,a1=S1=4,当n≥2时,an=Sn-Sn-1=n2+2n+1-[(n-1)2+2(n-1)+1]=2n+1,又a1=4不适合上式,∴an=4,   n=12n+1,  n≥2(II)∵1a1a2=14×5,当n≥2时,1anan+1=1(2n+1)(2n+3...