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文[1]-[4]都谈了递推数列a n=c·a n-1+d·bn通项公式的求法,笔者受益匪浅,深受启发.文[1]指出了形如a n=c·a n-1+d·bn(c≠0,c≠1,d≠0,b≠0)的递推关系式均可由a n+λ·bn=c·(a n-1+λ·bn-1构造等比数列处理;文[2]指出了文[1]的不足,当c=d时无法求出λ的值,因此,还应加上条件c≠b,并给出了当c=b时数列通项式的求法;文[3]用迭代法求递推数列a n=c·a n-1+
I [1] - [4] talk about the recursive sequence an = c · a n-1 + d · bn general formula of the formula, the author has benefited a lot, inspired by the text [1] = c · a n-1 + d · bn (c ≠ 0, c ≠ 1, d ≠ 0, b ≠ 0) can be expressed by a n + λ · bn = c · (a n-1 + λ · bn-1 construct the geometric sequence processing; [2] pointed out the deficiencies of [1], when c = d can not find the value of λ, therefore, we should also add the condition c ≠ b and give When c = b when the formula for the number of columns; text [3] iterative method for recursive sequence an = c · a n-1 +