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Question-101568




Question Number 101568 by Quvonchbek last updated on 03/Jul/20
Answered by mr W last updated on 03/Jul/20
x^(25) ×(((1+3)(1+3^2 )...(1+3^(25) ))/3^(1+2+...+25) )=1  x^(25) ×(((1+3)(1+3^2 )...(1+3^(25) ))/3^(13×25) )=1  ⇒x=(3^(13) /( (((1+3)(1+3^2 )...(1+3^(25) )))^(1/(25)) ))
$${x}^{\mathrm{25}} ×\frac{\left(\mathrm{1}+\mathrm{3}\right)\left(\mathrm{1}+\mathrm{3}^{\mathrm{2}} \right)…\left(\mathrm{1}+\mathrm{3}^{\mathrm{25}} \right)}{\mathrm{3}^{\mathrm{1}+\mathrm{2}+…+\mathrm{25}} }=\mathrm{1} \\ $$$${x}^{\mathrm{25}} ×\frac{\left(\mathrm{1}+\mathrm{3}\right)\left(\mathrm{1}+\mathrm{3}^{\mathrm{2}} \right)…\left(\mathrm{1}+\mathrm{3}^{\mathrm{25}} \right)}{\mathrm{3}^{\mathrm{13}×\mathrm{25}} }=\mathrm{1} \\ $$$$\Rightarrow{x}=\frac{\mathrm{3}^{\mathrm{13}} }{\:\sqrt[{\mathrm{25}}]{\left(\mathrm{1}+\mathrm{3}\right)\left(\mathrm{1}+\mathrm{3}^{\mathrm{2}} \right)…\left(\mathrm{1}+\mathrm{3}^{\mathrm{25}} \right)}} \\ $$

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