1-1-3-1-1-4-1-1-5-1-1-99-

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

(1 - \frac{1}{3}) (1 - \frac{1}{4}) (1 - \frac{1}{5}) \dots . (1 - \frac{1}{99}) = ?

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

let A_{n} = (1 - \frac{1}{3}) (1 - \frac{1}{4}) \dots . (1 - \frac{1}{n}) \Rightarrow A_{n} = \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times \dots \frac{n - 2}{n - 1} \times \frac{n - 1}{n}

\Rightarrow A_{n} = \frac{2}{n} and (1 - \frac{1}{3}) (1 - \frac{1}{4}) \dots (1 - \frac{1}{99}) = \frac{2}{99}

Answered by OlafThorendsen last updated on 20/Jul/20

P = \underset{k = 3}{\prod^{99}} (1 - \frac{1}{k})

P = \underset{k = 3}{\prod^{99}} \frac{k - 1}{k}

P = \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \dots . \frac{97}{98} \times \frac{98}{99}

P = \frac{2}{99}

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

\underset{n = 3}{\prod^{99}} (\frac{n}{n + 1}) = \frac{2}{3} . \frac{3}{4} . \frac{4}{5} \dots . \frac{98}{99} = 2 (\frac{1}{3} . \frac{3}{4} . \frac{4}{5} \dots . . \frac{97}{98} . \frac{98}{99}) = \frac{2}{99}