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
,求Tn.1
anan+1
答
(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...