求递推数列的通项,是高考数列综合题最为常见的考查内容之一,虽然试题立意“试验——猜测——证明”的思想,但抽象推演的方法,也可能有很好的通性,而且更为简捷.本文推介的就是这样一种方法,不妨统称为“待定系数法”.1 始作俑者:a_(n+1)=b·a_n+c若 b=1,则数列{a_n)是等差数列;若c=0,b≠0,则数列{a_n}是等比数列;若c≠0,b≠1,b≠0时呢?下面的解法是常用方法:设常数 k 是 c 分解所得,且满足:a_(n+1)-k=b It is one of the most common test items to determine the number of rehearsals. Although the test questions are based on the idea of ​​“test - guess - proof”, the abstract deduction method may also be very good Generality, but also more simple. This article is to promote such a method, may wish to collectively referred to as the “method of undetermined coefficient.” 1 Initiator: a_ (n +1) = b · a_n + c If b = 1, then the sequence {a_n) is the arithmetic sequence; if c = 0, b ≠ 0, then the sequence {a_n} is the geometric sequence; if c ≠ 0, b ≠ 1, b ≠ 0? The following solution is a common method: The constant k is obtained by decomposing c and satisfies: a_ (n + 1) -k = b