由递推公式求通项公式,历年都是高考考查的热点,也是学生学习的难点。有些类型的递推公式其实可以用待定系数法巧妙地转化为等差数列或等比数列实现轻松求解。现归纳如下:1 a_n=ka(n-1)+m(k≠1,m≠0)型该解法是先将其设为a_n-x=k(a_(n-1)-x),利用待定系数法求出x=m/(1-k),再用换元法转化为等比数列求通项。 By recursion formula to find the formula, over the years are hot spots for college entrance examination, but also the difficulty of student learning. In fact, some types of recursive formulas can be easily solved by the method of undetermined coefficients, which is subtly transformed into arithmetic sequences or geometric series. Now it is summarized as follows: 1 a_n = ka (n-1) + m (k ≠ 1, m ≠ 0) Type The solution is to first set it to a_n-x = k (a_ (n-1) -x) To be determined by the coefficient method to find x = m / (1-k), and then replaced by the metamorphic conversion method for the general term.