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题目:设各项均为正数的数列n a的前n项和为n S,已知1a1,且1 1()(1)n n n n Sa S a对一切nN*都成立。(1)若1,求数列n a的通项公式;(2)求的值,使数列n a是等差数列.该题是2014年苏、锡、常、镇四市3月份高三教学情况调查考试第19题.此题分值16分,而我们学生测试的结果得分仅3分左右.为什么该题的得分率如此之低本题条件中出现了n S与n a的递推关系式,按照常规思
Title: Let the first n terms and nS of the sequence n a, each of which is a positive number, be known as 1a1, and 1 1 (1) n n n n Sa S a holds for all nN *. (1) If 1, find the general formula of the sequence na; (2) Find the value so that the sequence na is the arithmetic sequence. This question is the 2014 senior high school teaching survey in March of the four cities of Jiangsu, Question 19. This question scores 16 points, and our students test results score only about 3. Why the question score is so low The problem conditions appear in the recursion n S and na relations, according to the conventional thinking