分式：计算1/(x-1)+1/(x-1)(x-2)+1/(x-2)(x-3)+...+1/(x-99)(x-100)
展开全部
思路：我们知道：1/3*2=(3-2)/3*2=3/3*2-2/3*2=1/2-1/3
类比得：1/n*(n-1)=[n-(n-1)]/n*(n-1)=n/n*(n-1)-(n-1)/n*(n-1)=1/(n-1)-1/n
所以1/(x-1)(x-2)=1/(x-2)-1/(x-1), 1/(x-99)(x-100)=1/(x-100)-1/(x-99), .....
所以1/(x-1)+ 1/(x-1)(x-2)+1/(x-2)(x-3)+.......+1/(x-99)(x-100)
=1/(x-1)+ 1/(x-2)-1/(x-1)+1/(x-3)-1/(x-2)+......+ 1/(x-100)-1/(x-99)
=1/(x-100)
类比得：1/n*(n-1)=[n-(n-1)]/n*(n-1)=n/n*(n-1)-(n-1)/n*(n-1)=1/(n-1)-1/n
所以1/(x-1)(x-2)=1/(x-2)-1/(x-1), 1/(x-99)(x-100)=1/(x-100)-1/(x-99), .....
所以1/(x-1)+ 1/(x-1)(x-2)+1/(x-2)(x-3)+.......+1/(x-99)(x-100)
=1/(x-1)+ 1/(x-2)-1/(x-1)+1/(x-3)-1/(x-2)+......+ 1/(x-100)-1/(x-99)
=1/(x-100)
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
