[tex]Notez~ \frac{1}{3}=a. \\ \\ 1+ \frac{1}{3}+ (\frac{1}{3})^2+ (\frac{1}{3})^3 +( \frac{1}{3} )^4= \\ \\ =1+a+a^2+a^3+a^4= \\ \\ = \frac{a ^{5}-1 }{a-1}= \\ \\ = \frac{ \frac{1}{243}-1 }{ \frac{1}{3}-1 }= \\ \\ = \frac{- \frac{242}{243} }{- \frac{2}{3} } = \\ \\ = \frac{242}{243}* \frac{3}{2}= \\ \\ = \frac{121}{81} .[/tex]
