2楼: Originally posted by 判官~三爷 at 2013-09-13 00:26:40
可以求助matlab

4楼: Originally posted by zjl20111015 at 2013-09-13 19:53:18
应该可以构造一个式子（x/3-1）^a*(x-1)^a的展开式中x^j的系数

5楼: Originally posted by 我要飞 at 2013-09-13 23:45:54
能详细告诉我吗？谢谢了...

6楼: Originally posted by hank612 at 2013-09-15 06:04:25
zjl20111015 的意思是说， 他可以算出楼主所求序列和的生成函数
\Sum_{j=0}^{\infty} （楼主的求和）* x^j = (1-x)^a * (1-x/3)^a.
这可以由 (1+x)^a = \Sum_{n=0}^{\infty} (a choose n) x^n 标准的Taylor 级数 ...

7楼: Originally posted by 我要飞 at 2013-09-15 07:59:06
能证明当j大于4时此式子恒正就可以了...