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-

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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

Solve the given equation :

(x + \frac{x}{3^{1}}) \cdot (x + \frac{x}{3^{2}}) \cdot (x + \frac{x}{3^{3}}) \cdot (x + \frac{x}{3^{4}}) \cdot \dots \cdot (x + \frac{x}{3^{25}}) = 1

A) \frac{3^{51}}{2 \cdot (3^{50} - 1)} B) \frac{3^{52}}{2 \cdot (3^{51} - 1)} C) \frac{3^{50}}{2 \cdot (3^{50} - 1)} D) \frac{3^{51}}{3^{51} - 1}

Please help with solution

Commented by abony1303 last updated on 09/Aug/20

Someone, pls help

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

a l l a n s w e r s a r e w r o n g .

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^{25} \times \frac{\underset{k = 1}{\prod^{25}} (1 + 3^{k})}{3^{\underset{k = 1}{\sum^{25}} k}} = 1

x^{25} = \frac{3^{25 \times 13}}{\underset{k = 1}{\prod^{25}} (1 + 3^{k})}

\Rightarrow x = \frac{3^{13}}{{[\underset{k = 1}{\prod^{25}} (1 + 3^{k})]}^{\frac{1}{25}}}

Commented by abony1303 last updated on 09/Aug/20

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

can we somehow find out the denominator

of the last 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

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