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Find-3-3-4-3-3-5-3-1-2-3-3-7-3-1-6-27-5-




Question Number 213589 by hardmath last updated on 09/Nov/24
Find:  (((3 - (3/4))∙(3 - (3/5))∙(3 - (1/2))∙(3 - (3/7))∙...∙(3 - (1/6)))/(27^5 )) = ?
$$\mathrm{Find}: \\ $$$$\frac{\left(\mathrm{3}\:-\:\frac{\mathrm{3}}{\mathrm{4}}\right)\centerdot\left(\mathrm{3}\:-\:\frac{\mathrm{3}}{\mathrm{5}}\right)\centerdot\left(\mathrm{3}\:-\:\frac{\mathrm{1}}{\mathrm{2}}\right)\centerdot\left(\mathrm{3}\:-\:\frac{\mathrm{3}}{\mathrm{7}}\right)\centerdot…\centerdot\left(\mathrm{3}\:-\:\frac{\mathrm{1}}{\mathrm{6}}\right)}{\mathrm{27}^{\mathrm{5}} }\:=\:? \\ $$
Answered by issac last updated on 09/Nov/24
((1/3))^(15) Π_(h=1) ^(15)  (3−(3/(h+3)))=(1/6)
$$\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{15}} \underset{{h}=\mathrm{1}} {\overset{\mathrm{15}} {\prod}}\:\left(\mathrm{3}−\frac{\mathrm{3}}{{h}+\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{6}} \\ $$
Commented by hardmath last updated on 09/Nov/24
  Thanks a lot, is there any other easy way to do this?
$$ \\ $$Thanks a lot, is there any other easy way to do this?
Answered by mehdee7396 last updated on 09/Nov/24
((3^(15) (1−(1/4))(1−(1/5))...(1−(1/(18))))/3^(15) )  =((3/4))((4/5))((5/6))...(((16)/(17)))(((17)/(18)))  =(1/6)
$$\frac{\mathrm{3}^{\mathrm{15}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}\right)…\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{18}}\right)}{\mathrm{3}^{\mathrm{15}} } \\ $$$$=\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\left(\frac{\mathrm{4}}{\mathrm{5}}\right)\left(\frac{\mathrm{5}}{\mathrm{6}}\right)…\left(\frac{\mathrm{16}}{\mathrm{17}}\right)\left(\frac{\mathrm{17}}{\mathrm{18}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\: \\ $$

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