计算

共有101项相加，有公因式a*b，提出公因式后发现这101项的分子均为0，分母均为相差1的两个数的乘积，由：文章源自玩技e族-https://www.playezu.com/249607.html

1/[a*(a+1)]可拆分成两项的差，即：(1/a)-(1/(a+1))。文章源自玩技e族-https://www.playezu.com/249607.html

所以原式可化为：文章源自玩技e族-https://www.playezu.com/249607.html

a*b*{[1/(a+b-1)-1/(a+b)]+[1/(a+b)-1/(a+b+1)]+…+[1/(a+b+99)-1/(a+b+100)]}大括号内，中间项可相消，只剩首尾两项，所以原式=a*b*[1/(a+b-1)-1/(a+b+100)]=101*a*b/[(a+b-1)*(a+b+100)]文章源自玩技e族-https://www.playezu.com/249607.html

ab[1/(a+b-1)(a+b)+1/(a+b)(a+b+1)+1/(a+b+1)(a+b+2)+…+1/(a+b+98)(a+b+99)+1/(a+b+99)(a+b+100)]=ab[1/(a+b-1)-1/(a+b)+1/(a+b)-1/(a+b+1)+1/(a+b+1)-1/(a+b+2)+…+1/(a+b+98)-1/(a+b+99)+1/(a+b+99)-1/(a+b+100)]文章源自玩技e族-https://www.playezu.com/249607.html

=ab[1/(a+b)-1/(a+b+100)]=75/49

知识问答