Question

Найдите сумму шести первых членов геометрической прогрессии (bn), в которой:

а)b1=27, q=1/3

б)b1=-9, q=2

в)b1=16, q=-1/2

г)b1=3√2, q=√2

Answer 1

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

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

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

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

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

Answer 2

спасибо