Question Number 201408 by Ari last updated on 05/Dec/23
Answered by mr W last updated on 05/Dec/23
$$=\frac{\mathrm{1}×\mathrm{3}}{\mathrm{2}^{\mathrm{2}} }×\frac{\mathrm{2}×\mathrm{4}}{\mathrm{3}^{\mathrm{2}} }×\frac{\mathrm{3}×\mathrm{5}}{\mathrm{4}^{\mathrm{2}} }×…×\frac{\mathrm{98}×\mathrm{100}}{\mathrm{99}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}×\cancel{\mathrm{3}}}{\mathrm{2}^{\cancel{\mathrm{2}}} }×\frac{\cancel{\mathrm{2}}×\cancel{\mathrm{4}}}{\cancel{\mathrm{3}^{\mathrm{2}} }}×\frac{\cancel{\mathrm{3}}×\cancel{\mathrm{5}}}{\cancel{\mathrm{4}^{\mathrm{2}} }}×…×\frac{\cancel{\mathrm{98}}×\mathrm{100}}{\mathrm{99}^{\cancel{\mathrm{2}}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{100}}{\mathrm{99}} \\ $$$$=\frac{\mathrm{50}}{\mathrm{99}} \\ $$
Answered by Mathspace last updated on 05/Dec/23
$$\prod_{{k}=\mathrm{2}} ^{{n}} \left(\mathrm{1}−\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right)={A}_{{n}} \\ $$$$\Rightarrow{A}_{{n}} =\prod_{{l}=\mathrm{2}} ^{{n}} \frac{\left({k}−\mathrm{1}\right)\left({k}+\mathrm{1}\right)}{{k}^{\mathrm{2}} } \\ $$$$=\prod_{{k}=\mathrm{2}} ^{{n}} \:\frac{{k}−\mathrm{1}}{{k}}×\frac{{k}+\mathrm{1}}{{k}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{4}}{\mathrm{3}}×…×\frac{{n}−\mathrm{2}}{{n}−\mathrm{1}}×\frac{{n}}{{n}−\mathrm{1}}× \\ $$$$\frac{{n}−\mathrm{1}}{{n}}×\frac{{n}+\mathrm{1}}{{n}}=\frac{\mathrm{1}}{\mathrm{2}}=\frac{{n}+\mathrm{1}}{\mathrm{2}{n}} \\ $$$${A}_{\mathrm{99}} =\frac{\mathrm{99}+\mathrm{1}}{\mathrm{2}×\mathrm{99}}=\frac{\mathrm{100}}{\mathrm{2}.\mathrm{99}}=\frac{\mathrm{50}}{\mathrm{99}} \\ $$
Answered by tri26112004 last updated on 06/Dec/23
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