一、问题的源出问题在于探讨“差比”数列(即一阶线性递推数列)通项公式的求法:若数列{an}满足a1=b,an+1=can+d(c≠1),其通项公式的求法一般采用如下的参数法,将递推数列转化为等比数列:设an+1+t=c(an+t),则an+1=can+(c-1)t, First, the problem of the source of the problem lies in the “difference ratio ” series (ie first order linear recursion sequence) general formula of the formula: if the sequence {an} satisfies a1 = b, an + 1 = can + d ≠ 1), the general formula of the general method of the following parameter method, the recursive sequence into the geometric series: an + 1 + t = c (an + t), then an + 1 = can + (c- 1) t,