Tentukan jumlah dari deret 81+27+9+...+1/3

a = 81
[tex]r = \frac{27}{81} = \frac{1}{3} \\ un = \frac{1}{3} \\ un = {ar}^{n - 1} \\ \frac{1}{ 3} = 81( { \frac{1}{3} }^{n - 1} ) \\ \frac{1}{243} = \frac{1}{3} {}^{n - 1} \\ ( { \frac{1}{3} }^{5} ) = \frac{1}{3} {}^{n - 1} \\ \\ n - 1 = 5 \\ n = 6[/tex]
[tex]s6 = \frac{81(1 - \frac{1}{3} {}^{6} )}{1 - \frac{1}{3} } \\ = 81( \frac{728}{729} ) \times \frac{3}{2} \\ = 81( \frac{364}{243} ) = 121 \frac{81}{243} = 121 \frac{1}{3} [/tex]
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