Question

Найдите сумму шести первых членов геометрической прогрессии (bn), в которой:а)b1=27, q=1/3б)b1=-9, q=2в)b1=16, q=-1/2г)b1=3√2, q=√2

Answer 1

Сумма первых n членов геометрической прогрессии:$S_n= \dfrac{b_1(q^n-1)}{q-1}$

$S_6= \dfrac{27\cdot(( \frac{1}{3} )^6-1)}{ \frac{1}{3} -1} =\dfrac{27\cdot( \frac{1}{729} -1)}{ \frac{1}{3} -1} =\dfrac{27\cdot( 1-\frac{1}{729} )}{ 1-\frac{1}{3} } =\dfrac{27\cdot \frac{728}{729} }{ \frac{2}{3} } = \\ = \dfrac{27\cdot 728\cdot3}{729\cdot 2} = \dfrac{364\cdot3}{27} = \dfrac{364}{9} =40 \dfrac{4}{9}$

$S_6= \dfrac{-9\cdot (2^6-1)}{ 2 -1} = \dfrac{-9\cdot (64-1)}{1} =-9\cdot 63=-567$

$S_6= \dfrac{16\cdot(( -\frac{1}{2} )^6-1)}{ -\frac{1}{2} -1} =\dfrac{16\cdot( \frac{1}{64} -1)}{ -\frac{1}{2} -1} =\dfrac{16\cdot( 1-\frac{1}{64} )}{ \frac{1}{2}+1 } =\dfrac{16\cdot \frac{63}{64} }{ \frac{3}{2} } = \\ = \dfrac{16\cdot 63\cdot2}{64\cdot 3} = \dfrac{21\cdot2}{4} = \dfrac{21}{2} =10.5$

$S_6= \dfrac{3 \sqrt{2} \cdot(( \sqrt{2} )^6-1)}{ \sqrt{2} -1} =\dfrac{3 \sqrt{2} \cdot( 8 -1)}{ \sqrt{2} -1} =\dfrac{3 \sqrt{2} \cdot7}{ \sqrt{2} -1 } =\dfrac{21 \sqrt{2} }{ \sqrt{2} -1 } = \\ =\dfrac{21 \sqrt{2}\cdot ( \sqrt{2} +1) }{ ( \sqrt{2} -1)( \sqrt{2} +1) } = \dfrac{21\cdot 2+21 \sqrt{2} }{ ( \sqrt{2} )^2-1^2 } = \dfrac{42+21 \sqrt{2} }{ 2-1 } =42+21 \sqrt{2}$