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一、关于一类数列通项的一种求法先看下面两个常见的例子。例1.已知数列{a_n}中a_0=1,a_1=1且a_n=a_(n-1)+2a_(n-2)(n=2,3,…) (1)求通项a_n·例2.已知数列{a_n}中a_0=0,a_1=1且a_n=3a_(n-1)-2a_(n-2)+1(n=2,3,…) (2)求通项a_n·为了解答上面提出的第一个问题,我们应用循环级数的方法。这种方法的主要点是以数列{a_n}中各项为系数组成一个幂级数,并取其前n项的和,用S_n表示,就是S_n=a_0+a_1x+a_2x~2+…+a_nx~n (3)由关系式(1)得 a_n-a_(n-1)-2a_(n-2)=0 (4)从而1.(a_nx~n)-x(a_(n-1)x~(n-1))-2x~2(a_(n-2)x~(n-2))=0 (5)(5)式说明:对于n≥2,依次以1,-x,-2x~2乘(3)两边,然后双方求其代数和,可以
First, a method of seeking a list of general terms First look at the following two common examples. Example 1. A_0=1, a_1=1, and a_n=a_(n-1)+2a_(n-2) (n=2,3,...) in the known sequence {a_n} (1) Finding term a_n· Example 2. A_0=0, a_1=1, and a_n=3a_(n-1)-2a_(n-2)+1(n=2,3,...) in the known sequence {a_n} (2) A_n· In order to answer the first question above, we apply the method of the loop series. The main point of this method is that the items in the series {a_n} constitute a power series and take the sum of the first n items, which is represented by S_n, that is S_n=a_0+a_1x+a_2x~2+...+a_nx ~n (3) From the relation (1) a_n-a_(n-1)-2a_(n-2)=0 (4) thus 1.(a_nx~n)-x(a_(n-1)x ~(n-1))-2x~2(a_(n-2)x~(n-2))=0 (5)(5) Note: For n≥2, 1, -x, - 2x~2 multiplications (3) on both sides, then both sides ask for their algebraic sum, can