已知数列{an}的前n项和Sn=+2n,Tn=1/(a1a2)+1/(a2a3)+1/(a3a4)+...+1/(ana
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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)+1/(a2a3)+1/(a3a4)+...+1/(ana](/uploads/image/z/2401982-62-2.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%3D%2B2n%2CTn%3D1%2F%28a1a2%29%2B1%2F%28a2a3%29%2B1%2F%28a3a4%29%2B...%2B1%2F%28ana)
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)
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)
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