Question Number 104246 by Anindita last updated on 20/Jul/20

(1−(1/3))(1−(1/4))(1−(1/5))....(1−(1/(99))) =?

Answered by mathmax by abdo last updated on 20/Jul/20

let A_n =(1−(1/3))(1−(1/4))....(1−(1/n)) ⇒A_n =(2/3)×(3/4)×(4/5)×...((n−2)/(n−1))×((n−1)/n)  ⇒A_n =(2/n) and (1−(1/3))(1−(1/4))...(1−(1/(99))) =(2/(99))

Answered by OlafThorendsen last updated on 20/Jul/20

P = Π_(k=3) ^(99) (1−(1/k))  P = Π_(k=3) ^(99) ((k−1)/k)  P = (2/3)×(3/4)×(4/5)....((97)/(98))×((98)/(99))  P = (2/(99))

Answered by Dwaipayan Shikari last updated on 20/Jul/20

Π_(n=3) ^(99) ((n/(n+1)))=(2/3).(3/4).(4/5)....((98)/(99))=2((1/3).(3/4).(4/5).....((97)/(98)).((98)/(99)))=(2/(99))