已知数列{an}的前n项和Sn=+2n,Tn=1/(a1a2)+1/(a2a3)+1/(a3a4)+...+1/(anan+1),求Tn

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已知数列{an}的前n项和Sn=+2n,Tn=1/(a1a2)+1/(a2a3)+1/(a3a4)+...+1/(anan+1),求Tn已知数列{an}的前n项和Sn=+2n,Tn=1/(a1a2)+

已知数列{an}的前n项和Sn=+2n,Tn=1/(a1a2)+1/(a2a3)+1/(a3a4)+...+1/(anan+1),求Tn
已知数列{an}的前n项和Sn=+2n,Tn=1/(a1a2)+1/(a2a3)+1/(a3a4)+...+1/(anan+1),求Tn

已知数列{an}的前n项和Sn=+2n,Tn=1/(a1a2)+1/(a2a3)+1/(a3a4)+...+1/(anan+1),求Tn
Sn=n²+2n是吧.
n=1时,a1=S1=1²+2×1=3
n≥2时,an=Sn-S(n-1)=n²+2n-[(n-1)²+2(n-1)]=2n+1
n=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+1
1/[ana(n+1)]=1/[(2n+1)(2(n+1)+1)]=(1/2)[1/(2n+1)-1/(2(n+1)+1)]
Tn=1/(a1a2)+1/(a2a3)+...+1/[ana(n+1)]
=(1/2)[1/3-1/5+1/5-1/7+...+1/(2n+1)-1/(2(n+1)+1)]
=(1/2)[1/3-1/(2n+3)]
=n/(6n+9)

Sn=n^2+2n
an=Sn-S(n-1)
=n^2+2n-[(n-1)2+2(n-1)]
=2n+1
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)]

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