在数列题中,从递推公式出发求数列的通项公式的题目,题型繁杂,方法琐碎,这类题也是高考的一个难点。解题时重点要分清不同的模型。用不同的对策。通过等价变形运用化归思想,化为递推数列的六种基本模型之一,再转化为等差或等比数列,从而达到熟练解决问题的目的。一、a_(n+1)=a_n+f(n)型例1在数列{a_n}中,a_1=1,a_(n+1)=a_n+1/n(n+1)(n∈N~*),求a_n。解:由a_(n+1)-a_n=1/n(n+1)得: In the series of questions, starting from the recursive formula to solve the problem of the general formula of the series, the questions are complex, the method is trivial, such questions are also a difficult college entrance examination. Solve the problem to distinguish between different models. Use different strategies. Through the use of equivalence transformation, the idea of ​​the return to normalization is transformed into one of the six basic models of recursion sequences and then transformed into an equal-difference or an equal-number sequence so as to achieve the purpose of skillfully solving the problem. Example 1 In the sequence {a_n}, a_1 = 1, a_ (n + 1) = a_n + 1 / n (n + 1) (n∈N ~ *), Find a_n. Solution: From a_ (n + 1) -a_n = 1 / n (n + 1):