最近,广州市几个区联合举行了调研考试(即期中区统考),我们也参加了这次考试,有一道数列解答题的第三问,我们区没有一人做对(听教师说).题目如下:在数列{an}中,a1=1,a2=3,an+2=3an+1-kan(k≠0)对任意n∈N*成立,令bn=an+1-an,且{bn}是等比数列.(1)求实数k的值;(2)求数列{an}的通项公式; Recently, several districts in Guangzhou jointly held the research examination (ie, the Central District Unified Test). We also participated in the examination. There was a third question in the series and none in our district was correct. Is as follows: a1 = 1, a2 = 3, an + 2 = 3an + 1-kan (k ≠ 0) holds for any n∈N * in the sequence {an} such that bn = an + 1-an and { bn} is the geometric sequence. (1) Find the value of real number k; (2) Find the general formula of {an}