问题:已知数列{a_n}满足a_1=1,a_(n+1)~2=9a_n~2(n∈N~*).(1)定义c_n=|a_n|-k|a_(n+1)|(n∈N~*).设数列{|a_n|},{c_n}的前n项和分别为T_n,S_n.如果对于所有的正整数n,不等式S_n>knT_n都成立,求k的取值范围.(2)若数列{a_n+2·3~n}是等比数列,求数列{a_n}的通项公式.这是2011年的一道自主招生数学题,题目并不难,但得分率极低.巧妙地考查出了考生在思维能力 Problem: The known sequence {a_n} satisfies a_1 = 1, a_ (n + 1) ~ 2 = 9a_n ~ 2 (n∈N ~ *). (1) Definition c_n = | a_n | -k | a_ (n + 1 (n ∈ N ~ *). Let the first n terms of the {n - N} and {c_n} be T_n and S_n respectively. If the inequality S_n> knT_n holds for all positive integers n, (2) If the sequence {a_n + 2 · 3 ~ n} is a geometric sequence, find the general formula of {a_n}. This is a self-enrollment mathematics problem in 2011, the title is not difficult, but Very low scoring rate cleverly test the candidates in thinking ability