Question Number 107198 by abony1303 last updated on 09/Aug/20
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation}: \\ $$$$\left(\mathrm{x}+\frac{\mathrm{x}}{\mathrm{3}^{\mathrm{1}} }\right)\centerdot\left(\mathrm{x}+\frac{\mathrm{x}}{\mathrm{3}^{\mathrm{2}} }\right)\centerdot\left(\mathrm{x}+\frac{\mathrm{x}}{\mathrm{3}^{\mathrm{3}} }\right)\centerdot\left(\mathrm{x}+\frac{\mathrm{x}}{\mathrm{3}^{\mathrm{4}} }\right)\centerdot…\centerdot\left(\mathrm{x}+\frac{\mathrm{x}}{\mathrm{3}^{\mathrm{25}} }\right)=\mathrm{1} \\ $$$$\left.\mathrm{A}\left.\right)\left.\frac{\mathrm{3}^{\mathrm{51}} }{\mathrm{2}\centerdot\left(\mathrm{3}^{\mathrm{50}} −\mathrm{1}\right)}\left.\:\:\mathrm{B}\right)\frac{\mathrm{3}^{\mathrm{52}} }{\mathrm{2}\centerdot\left(\mathrm{3}^{\mathrm{51}} −\mathrm{1}\right)}\:\:\mathrm{C}\right)\:\frac{\mathrm{3}^{\mathrm{50}} }{\mathrm{2}\centerdot\left(\mathrm{3}^{\mathrm{50}} −\mathrm{1}\right)}\:\:\mathrm{D}\right)\:\frac{\mathrm{3}^{\mathrm{51}} }{\mathrm{3}^{\mathrm{51}} −\mathrm{1}} \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{with}\:\mathrm{solution} \\ $$
Commented by abony1303 last updated on 09/Aug/20
$$\mathrm{Someone},\:\mathrm{pls}\:\mathrm{help} \\ $$
Commented by mr W last updated on 09/Aug/20
$${all}\:{answers}\:{are}\:{wrong}. \\ $$
Answered by mr W last updated on 09/Aug/20
![x^(25) ×((Π_(k=1) ^(25) (1+3^k ))/3^(Σ_(k=1) ^(25) k) )=1 x^(25) =(3^(25×13) /(Π_(k=1) ^(25) (1+3^k ))) ⇒x=(3^(13) /([Π_(k=1) ^(25) (1+3^k )]^(1/(25)) ))](https://www.tinkutara.com/question/Q107224.png)
$${x}^{\mathrm{25}} ×\frac{\underset{{k}=\mathrm{1}} {\overset{\mathrm{25}} {\prod}}\left(\mathrm{1}+\mathrm{3}^{{k}} \right)}{\mathrm{3}^{\underset{{k}=\mathrm{1}} {\overset{\mathrm{25}} {\sum}}{k}} }=\mathrm{1} \\ $$$${x}^{\mathrm{25}} =\frac{\mathrm{3}^{\mathrm{25}×\mathrm{13}} }{\underset{{k}=\mathrm{1}} {\overset{\mathrm{25}} {\prod}}\left(\mathrm{1}+\mathrm{3}^{{k}} \right)} \\ $$$$\Rightarrow{x}=\frac{\mathrm{3}^{\mathrm{13}} }{\left[\underset{{k}=\mathrm{1}} {\overset{\mathrm{25}} {\prod}}\left(\mathrm{1}+\mathrm{3}^{{k}} \right)\right]^{\frac{\mathrm{1}}{\mathrm{25}}} } \\ $$
Commented by abony1303 last updated on 09/Aug/20
$$\mathrm{can}\:\mathrm{we}\:\mathrm{somehow}\:\mathrm{find}\:\mathrm{out}\:\mathrm{the}\:\mathrm{denominator} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{last}\:\mathrm{row}? \\ $$
Commented by hgrocks last updated on 09/Aug/20
No That's not easy at all , the denominator is nothing but value of function with roots -3,-6,-9..... at x = -1 and finding that function is not simple