В геометрической прогрессии b_1=1/81; q=3. Найдите b_6; S_6.

\[b_{1} = \frac{1}{81};\ \ q = 3:\]

\[b_{6} = b_{1} \cdot q^{5} = \left( \frac{1}{3} \right)^{4} \cdot 3^{5} = 3.\]

\[S_{6} = \frac{b_{1}\left( q^{n} - 1 \right)}{q - 1} =\]

\[= \frac{\frac{1}{81} \cdot \left( \left( 3^{6} \right) - 1 \right)}{3 - 1} = \frac{\frac{1}{81} \cdot 728}{2} =\]

\[= \frac{364}{81} = 4\frac{40}{81}.\]

\[Ответ:3;\ \ 4\frac{40}{81}.\]