2011年广东高考数学第20题第(1)问是:设b>0,数列{an}满足a1=b,an=(nan-1)/(an-1+2n-1)(n≥2),求数列{an}的通项公式.看到这个问题,使我们想起了2006年江西高考22题第(1)问:已知数列{an}满足:a1=32,且an=(3nan-1)/(2an-1+n-1)(n≥2,n∈N*),求数列{an}的通项公式.虽然考试的时间不同,但考生还是感 (1) of the math test of Guangdong college entrance examination in 2011 is: Suppose b> 0, the sequence {an} satisfies a1 = b, an = (nan- 1) / (an- 1 + 2n- 1) ), The number of columns {an} general formula to see this question, reminds us of the 2006 Jiangxi college entrance examination 22 questions (1) Q: known sequence {an} meet: a1 = 32, and an = (3nan -1) / (2an-1 + n-1) (n≥2, n∈N *), find the general formula of {an} .Although the examination time is different, but the candidates still feel