Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 107198 by abony1303 last updated on 09/Aug/20

Solve the given equation: (x+(x/3^1 ))∙(x+(x/3^2 ))∙(x+(x/3^3 ))∙(x+(x/3^4 ))∙...∙(x+(x/3^(25) ))=1 A)(3^(51) /(2∙(3^(50) −1))) B)(3^(52) /(2∙(3^(51) −1))) C) (3^(50) /(2∙(3^(50) −1))) D) (3^(51) /(3^(51) −1)) Please help with solution

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation}: \\ $$$$\left(\mathrm{x}+\frac{\mathrm{x}}{\mathrm{3}^{\mathrm{1}} }\right)\centerdot\left(\mathrm{x}+\frac{\mathrm{x}}{\mathrm{3}^{\mathrm{2}} }\right)\centerdot\left(\mathrm{x}+\frac{\mathrm{x}}{\mathrm{3}^{\mathrm{3}} }\right)\centerdot\left(\mathrm{x}+\frac{\mathrm{x}}{\mathrm{3}^{\mathrm{4}} }\right)\centerdot...\centerdot\left(\mathrm{x}+\frac{\mathrm{x}}{\mathrm{3}^{\mathrm{25}} }\right)=\mathrm{1} \\ $$$$\left.\mathrm{A}\left.\right)\left.\frac{\mathrm{3}^{\mathrm{51}} }{\mathrm{2}\centerdot\left(\mathrm{3}^{\mathrm{50}} −\mathrm{1}\right)}\left.\:\:\mathrm{B}\right)\frac{\mathrm{3}^{\mathrm{52}} }{\mathrm{2}\centerdot\left(\mathrm{3}^{\mathrm{51}} −\mathrm{1}\right)}\:\:\mathrm{C}\right)\:\frac{\mathrm{3}^{\mathrm{50}} }{\mathrm{2}\centerdot\left(\mathrm{3}^{\mathrm{50}} −\mathrm{1}\right)}\:\:\mathrm{D}\right)\:\frac{\mathrm{3}^{\mathrm{51}} }{\mathrm{3}^{\mathrm{51}} −\mathrm{1}} \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{with}\:\mathrm{solution} \\ $$

Commented by abony1303 last updated on 09/Aug/20

Someone, pls help

$$\mathrm{Someone},\:\mathrm{pls}\:\mathrm{help} \\ $$

Commented by mr W last updated on 09/Aug/20

all answers are wrong.

$${all}\:{answers}\:{are}\:{wrong}. \\ $$

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)) ))

$${x}^{\mathrm{25}} ×\frac{\underset{{k}=\mathrm{1}} {\overset{\mathrm{25}} {\prod}}\left(\mathrm{1}+\mathrm{3}^{{k}} \right)}{\mathrm{3}^{\underset{{k}=\mathrm{1}} {\overset{\mathrm{25}} {\sum}}{k}} }=\mathrm{1} \\ $$$${x}^{\mathrm{25}} =\frac{\mathrm{3}^{\mathrm{25}×\mathrm{13}} }{\underset{{k}=\mathrm{1}} {\overset{\mathrm{25}} {\prod}}\left(\mathrm{1}+\mathrm{3}^{{k}} \right)} \\ $$$$\Rightarrow{x}=\frac{\mathrm{3}^{\mathrm{13}} }{\left[\underset{{k}=\mathrm{1}} {\overset{\mathrm{25}} {\prod}}\left(\mathrm{1}+\mathrm{3}^{{k}} \right)\right]^{\frac{\mathrm{1}}{\mathrm{25}}} } \\ $$

Commented by abony1303 last updated on 09/Aug/20

can we somehow find out the denominator of the last row?

$$\mathrm{can}\:\mathrm{we}\:\mathrm{somehow}\:\mathrm{find}\:\mathrm{out}\:\mathrm{the}\:\mathrm{denominator} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{last}\:\mathrm{row}? \\ $$

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com