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(2008年浙江理22题)已知数列{a_n},a_n≥0,a_1=0,a_(n+1)~2+a_(n+1)-1=a_n~2(n∈N~*)。记:S_n=a_1+a_2+…+a_n,T_n=1/(1+a_1)+1/((1+a_1)(1+a_2))+…+1/((1+a_1)(1+a_2)(1+a_n))。求证:当n∈N~*时,(Ⅰ)a_nn-2;(Ⅲ)T_n<3。以数列递推式为给出条件的综合题,除了求出通项公式(本题就不能求通项公式)外,解题的突破口只能对递推式进行分析、
(N = 0), a_n (n + 1) ~ 2 + a_ (n + 1) -1 = a_n ~ 2 (n∈N ~ *) ). Note: S_n = a_1 + a_2 + ... + a_n, T_n = 1 / (1 + a_1) +1 / (1 + a_1) (1 + a_2)) + ... + 1 / (1 + a_1) ) (1 + a_n)). Proof: When n∈N ~ *, (Ⅰ) a_n n-2; (Ⅲ) T_n <3. To sequence recursion to give the conditions of the comprehensive question, in addition to the general formula (this question can not find the formula), the problem-solving breakthrough can only be recursive analysis,