论文部分内容阅读
数列求和的方法很多,己有许多杂志刊登了各种数列求和方法的文章,本文提及的循环求和法,其思想方法是通过式子变形,使所求和重复出现,造成循环,亦即构造出含有所求和S的方程S=f(s),然后解出S。问题:求 sum from k=1 to n (k·2~k)sum from k=1 to n (k·2~k)=sum from k=0 to (n-1) ((k+1)2~(k+1))=2 sum from k=0 to (n-1) k2~k+sum from k= to (n-1) (2(k+1))=2[sum from k=1 to n (k·2~k-n·2~n)]+sum from k=1 to n 2~k∴ sum from k=1 to n (k·2~k)=n·2~(n+1)-(2~(n+1)-2) 有许多同志会感兴趣于研究sum from k=1 to n (k~p 2~k)
There are many methods for summation of numbers, and many magazines have published articles on various methods for summation of numbers. The method of circular summation referred to in this paper is based on the transformation of formulae so that the sum is repeated and the circulation is created. That is, the equation S=f(s) containing the sum S is constructed, and then S is solved. Problem: Find sum from k=1 to n (k·2~k)sum from k=1 to n (k·2~k)=sum from k=0 to (n-1) ((k+1)2 ~(k+1))=2 sum from k=0 to (n-1) k2~k+sum from k= to (n-1) (2(k+1))=2[sum from k=1 To n (k·2~kn·2~n)]+sum from k=1 to n 2~k∴ sum from k=1 to n (k·2~k)=n·2~(n+1) -(2~(n+1)-2) There are many comrades who are interested in the study sum from k=1 to n (k~p 2~k)