已知数列{an}的前n项和Sn=+2n,Tn=1/(a1a2)+1/(a2a3)+1/(a3a4)+...+1/(anan+1),求Tn
问题描述：
已知数列{an}的前n项和Sn=+2n,Tn=1/(a1a2)+1/(a2a3)+1/(a3a4)+...+1/(anan+1),求Tn
答
Sn=n²+2n是吧.n=1时,a1=S1=1²+2×1=3n≥2时,an=Sn-S(n-1)=n²+2n-[(n-1)²+2(n-1)]=2n+1n=1时,a1=2×1+1=3,同样满足通项公式数列{an}的通项公式为an=2n+11/[ana(n+1)]=1/[(2n+1)(2(n+1)+1)]=(1/2)...