若数列{An}满足A1=1,A(n+1)=An/(2An + 1)
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若数列{An}满足A1=1,A(n+1)=An/(2An + 1)
(n∈N+)
(1)求A2,A3
(2)判断数列{1/An}是否成等差数列,并说明理由.
(3)求证:1/3≤A1A2+A2A3+A3A4+.+AnA(n+1)
(n∈N+)
(1)求A2,A3
(2)判断数列{1/An}是否成等差数列,并说明理由.
(3)求证:1/3≤A1A2+A2A3+A3A4+.+AnA(n+1)
1) 1/3,1/5
2)倒数变换一下即可证明
从该步骤得到an=1/(2n-1)
3)
T=(1/1*1/3+1/3*1/5+1/5*1/7+……+[1/(2n-3)][1/(2n-1)]
=1/2(1-1/3+1/3-1/5+1/5-1/7+……+1/(2n-5)-1/(2n-3)+1/(2n-3)-1/(2n-1)
=1/2(1-1/(2n-1)
=n/(2n-1) ------下面引用
=1 所以 n-1>=0
原式n/(2n-1)>=n/[(2n-1)-(n-1)]
=n/3n=1/3【左边式子得证】
2)倒数变换一下即可证明
从该步骤得到an=1/(2n-1)
3)
T=(1/1*1/3+1/3*1/5+1/5*1/7+……+[1/(2n-3)][1/(2n-1)]
=1/2(1-1/3+1/3-1/5+1/5-1/7+……+1/(2n-5)-1/(2n-3)+1/(2n-3)-1/(2n-1)
=1/2(1-1/(2n-1)
=n/(2n-1) ------下面引用
=1 所以 n-1>=0
原式n/(2n-1)>=n/[(2n-1)-(n-1)]
=n/3n=1/3【左边式子得证】
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