由数列递推公式(相邻项之间的等式),求数列的通项是数列一章的热点难点,除了“试验一猜想”这种特殊的处理方法外,我们还应当掌握抽象的推理演算,以适用通性,通法,为此本文就如何由数列递推公式求数列通项问题作以小结.一、型如an+1=λan+c(λ≠1,c≠0)例1已知数列{an}有a1=1,an+1=2an+1,求通项an.解:因为an+1=2an+1,设它可变形为an+1+x=2(an+x), From the serial recursion formula (the equation between adjacent terms), the general term for the column of numbers is a hot and difficult point in the chapter of sequence. In addition to the special treatment of “test-and-guess”, we should also grasp the abstract To solve this problem, this paper summarizes how to solve the general term problem by using the sequence recursion formula. For example, an + 1 = λan + c (λ ≠ 1, c ≠ 0 Example 1 It is known that the sequence {a} has a1 = 1, an + 1 = 2an + 1, and finds the general term an. Solution: Let an + 1 = 2an + 1 denote it as an + 1 + x = 2 (an + x),