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Question Number 12326 by sin (x) last updated on 19/Apr/17

Π_(n=1) ^x a_n =9^(x!)   a_4 =?

$$\underset{{n}=\mathrm{1}} {\overset{{x}} {\prod}}{a}_{{n}} =\mathrm{9}^{{x}!} \\ $$$${a}_{\mathrm{4}} =? \\ $$

Answered by nume1114 last updated on 19/Apr/17

a_x =((a_1 ×a_2 ×...×a_(x−1) ×a_x )/(a_1 ×a_2 ×...×a_(x−1)         ))       =((Π_(n=1) ^x a_n )/(Π_(n=1) ^(x−1) a_n ))=(9^(x!) /9^((x−1)!) )=9^(x!−(x−1)!)   a_4 =9^(4!−(4−1)!) =9^(24−6) =9^(18)

$${a}_{{x}} =\frac{{a}_{\mathrm{1}} ×{a}_{\mathrm{2}} ×...×{a}_{{x}−\mathrm{1}} ×{a}_{{x}} }{{a}_{\mathrm{1}} ×{a}_{\mathrm{2}} ×...×{a}_{{x}−\mathrm{1}} \:\:\:\:\:\:\:\:} \\ $$$$\:\:\:\:\:=\frac{\underset{{n}=\mathrm{1}} {\overset{{x}} {\prod}}{a}_{{n}} }{\underset{{n}=\mathrm{1}} {\overset{{x}−\mathrm{1}} {\prod}}{a}_{{n}} }=\frac{\mathrm{9}^{{x}!} }{\mathrm{9}^{\left({x}−\mathrm{1}\right)!} }=\mathrm{9}^{{x}!−\left({x}−\mathrm{1}\right)!} \\ $$$${a}_{\mathrm{4}} =\mathrm{9}^{\mathrm{4}!−\left(\mathrm{4}−\mathrm{1}\right)!} =\mathrm{9}^{\mathrm{24}−\mathrm{6}} =\mathrm{9}^{\mathrm{18}} \\ $$

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