搜狗做题网已知数列{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
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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