Question

exercicio pg infinita
$x^{3} + \frac{ x^{3}}{3} + \frac{ x^{3}}{9} + ... = \frac{81}{16}$

Answer 1

q = a2/a1  
q =( x³/3 )/(x³/1) =  x³/3 * 1/x³ = 1/3 ***
S = a1/ ( 1 - q)
S=  x³ / ( 1/1 - 1/3)
S = x³/(  2/3 )
x³ / ( 2/3) = 81/16
1( x³ ) = 2/3( 81/16)
1/x³ = 162/48
x³ = 162/48  = 27/8 ***= 3³ / 2²   = ( 3/2)³ 

Answer 2

$\\ x^3 + \frac{x^3}{3} + \frac{x^3}{9} + ... = \frac{81}{16} \\\\ x^3 \cdot \underbrace{(1 + \frac{1}{3} + \frac{1}{9} + ...)}_{S_n} = \frac{81}{16}$

 Encontremos a soma dos termos da P.G em destaque:

- quanto à razão, temos:

$\\ q = \frac{a_2}{a_1} \\\\ q = \frac{1}{3} \div 1 \\\\ \boxed{q = \frac{1}{3}}$

- quanto à soma, teremos:

$\\ S_n = \frac{a_1}{1 - q} \\\\ S_n = \frac{1}{1 - \frac{1}{3}} \\\\ S_n = 1 \div \frac{3 - 1}{3} \\\\ S_n = 1 \times \frac{3}{3 - 1} \\\\ \boxed{S_n = \frac{3}{2}}$

 Por fim, concluímos que:

$\\x^3\cdot\underbrace{(1+\frac{1}{3}+\frac{1}{9}+...)}_{S_n}=\frac{81}{16} \\\\ x^3 \cdot \frac{3}{2} = \frac{81}{16} \\\\ x^3 = \frac{27}{8} \\\\ x = \sqrt[3]{\frac{3^3}{2^3}} \\\\ \boxed{\boxed{x = \frac{3}{2}}}$