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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=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文章源自玩技e族-https://www.playezu.com/249607.html

知识问答文章源自玩技e族-https://www.playezu.com/249607.html