Question

Ajutor :) .....................

Answer 1

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Answer 2

A doua paranteză are suma termenilor unei progresii geometrice cu rația q= 1/3.

$\it S_n = b_1\cdot\dfrac{q^n-1}{q-1}$

În cazul problemei date, avem:


[tex]\it S_{2004} = 1\cdot\dfrac{\left(\dfrac{1}{3}\right)^{2004}-1}{\dfrac{1}{3}-1} =\dfrac{1-\dfrac{1}{3^{2004}}}{1-\dfrac{1}{3}} =\dfrac{\dfrac{3^{2004}-1}{3^{2004}}}{\dfrac{2}{3}} = \\\;\\ \\\;\\ =\dfrac{3^{2004}-1}{3^{2004}}\cdot\dfrac{3}{2} =\dfrac{3^{2004}-1}{2\cdot3^{2003}} [/tex]


Acum, ecuația devine:



[tex]\it x\cdot3^{2004} = (3^{2004}-1):\dfrac{3^{2004}-1}{2\cdot3^{2003}} \\\;\\ \\\;\\ Notez\ \ 3^{2003} = a \Rightarrow 3^{2004} = 3a, rezult\breve{a}: \\\;\\ \\\;\\ x\cdot3a = (3a-1) : \dfrac{3a-1}{2a} \Rightarrow x\cdot3a = (3a-1) \cdot \dfrac{2a}{3a-1} \Rightarrow \\\;\\ \\\;\\ \Rightarrow x\cdot3a=2a \Rightarrow x = \dfrac{2a}{3a} \Rightarrow x = \dfrac{2}{3}. [/tex]