摘要:②记Tn=a1+a2+a3+a4+-+a+ a.求证:Tn=4n-1+Tn-1
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已知数列{an},an≥0,a1=0,an+12+an+1-1=an2(n∈N•).记Sn=a1+a2+…+an.Tn=
+
+…+
.
求证:当n∈N•时,
(Ⅰ)an<an+1;
(Ⅱ)Sn>n-2. 查看习题详情和答案>>
1 |
1+a1 |
1 |
(1+a1)(1+a2) |
1 |
(1+a1)(1+a2)…(1+an) |
求证:当n∈N•时,
(Ⅰ)an<an+1;
(Ⅱ)Sn>n-2. 查看习题详情和答案>>