Ответ: спасибо
Ответ: Сумма первых n членов геометрической прогрессии: [tex]S_n= \dfrac{b_1(q^n-1)}{q-1} [/tex][tex]S_6= \dfrac{27\cdot(( \frac{1}{3} )^6-1)}{ \frac{1}{3} -1} =\dfrac{27\cdot( \frac{1}{729} -1)}{ \frac{1}{3} -1} =\dfrac{27\cdot( 1-\frac{1}{729} )}{ 1-\frac{1}{3} } =\dfrac{27\cdot \frac{728}{729} }{ \frac{2}{3} } = \\ = \dfrac{27\cdot 728\cdot3}{729\cdot 2} = \dfrac{364\cdot3}{27} = \dfrac{364}{9} =40 \dfrac{4}{9} [/tex][tex]S_6= \dfrac{-9\cdot (2^6-1)}{ 2 -1} = \dfrac{-9\cdot (64-1)}{1} =-9\cdot 63=-567[/tex][tex]S_6= \dfrac{16\cdot(( -\frac{1}{2} )^6-1)}{ -\frac{1}{2} -1} =\dfrac{16\cdot( \frac{1}{64} -1)}{ -\frac{1}{2} -1} =\dfrac{16\cdot( 1-\frac{1}{64} )}{ \frac{1}{2}+1 } =\dfrac{16\cdot \frac{63}{64} }{ \frac{3}{2} } = \\ = \dfrac{16\cdot 63\cdot2}{64\cdot 3} = \dfrac{21\cdot2}{4} = \dfrac{21}{2} =10.5[/tex][tex]S_6= \dfrac{3 \sqrt{2} \cdot(( \sqrt{2} )^6-1)}{ \sqrt{2} -1} =\dfrac{3 \sqrt{2} \cdot( 8 -1)}{ \sqrt{2} -1} =\dfrac{3 \sqrt{2} \cdot7}{ \sqrt{2} -1 } =\dfrac{21 \sqrt{2} }{ \sqrt{2} -1 } = \\ =\dfrac{21 \sqrt{2}\cdot ( \sqrt{2} +1) }{ ( \sqrt{2} -1)( \sqrt{2} +1) } =
\dfrac{21\cdot 2+21 \sqrt{2} }{ ( \sqrt{2} )^2-1^2 } =
\dfrac{42+21 \sqrt{2} }{ 2-1 } =42+21 \sqrt{2} [/tex]