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

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

$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}\right)….\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{99}}\right)\:=? \\ $$

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))

$$\mathrm{let}\:\mathrm{A}_{\mathrm{n}} =\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)….\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{n}}\right)\:\Rightarrow\mathrm{A}_{\mathrm{n}} =\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{4}}{\mathrm{5}}×…\frac{\mathrm{n}−\mathrm{2}}{\mathrm{n}−\mathrm{1}}×\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}} \\ $$$$\Rightarrow\mathrm{A}_{\mathrm{n}} =\frac{\mathrm{2}}{\mathrm{n}}\:\mathrm{and}\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)…\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{99}}\right)\:=\frac{\mathrm{2}}{\mathrm{99}} \\ $$

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))

$$\mathrm{P}\:=\:\underset{{k}=\mathrm{3}} {\overset{\mathrm{99}} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{{k}}\right) \\ $$$$\mathrm{P}\:=\:\underset{{k}=\mathrm{3}} {\overset{\mathrm{99}} {\prod}}\frac{{k}−\mathrm{1}}{{k}} \\ $$$$\mathrm{P}\:=\:\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{4}}{\mathrm{5}}….\frac{\mathrm{97}}{\mathrm{98}}×\frac{\mathrm{98}}{\mathrm{99}} \\ $$$$\mathrm{P}\:=\:\frac{\mathrm{2}}{\mathrm{99}} \\ $$

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))

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

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