Respuesta :

[tex]\displaystyle 1+ \frac{1}{3} + \frac{1}{9} + \frac{1}{27} = \frac{1}{3^0}+ \frac{1}{3^1}+ \frac{1}{3^2} + \frac{1}{3^3} =\left( \frac{1}{3}\right)^0+\left( \frac{1}{3}\right)^1+\left( \frac{1}{3}\right)^2+\\\\\left( \frac{1}{3}\right)^3=\sum_{x=0}^3\left( \frac{1}{3}\right)^x \, \text{or }\, \sum_{i=1}^4\left( \frac{1}{3}\right)^{i-1}[/tex]

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