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

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

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

$$\mathrm{let}{\mathrm{A}}_{\mathrm{n}}=(1-\frac{1}{3})(1-\frac{1}{4})....(1-\frac{1}{\mathrm{n}})\Rightarrow {\mathrm{A}}_{\mathrm{n}}=\frac{2}{3}\times \frac{3}{4}\times \frac{4}{5}\times ...\frac{\mathrm{n}-2}{\mathrm{n}-1}\times \frac{\mathrm{n}-1}{\mathrm{n}}$$$$\Rightarrow {\mathrm{A}}_{\mathrm{n}}=\frac{2}{\mathrm{n}}\mathrm{and}(1-\frac{1}{3})(1-\frac{1}{4})...(1-\frac{1}{99})=\frac{2}{99}$$

Answered by OlafThorendsen last updated on 20/Jul/20

$$\mathrm{P}=\underset{k=3}{\prod ^{99}}(1-\frac{1}{k})$$$$\mathrm{P}=\underset{k=3}{\prod ^{99}}\frac{k-1}{k}$$$$\mathrm{P}=\frac{2}{3}\times \frac{3}{4}\times \frac{4}{5}....\frac{97}{98}\times \frac{98}{99}$$$$\mathrm{P}=\frac{2}{99}$$

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

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

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