Question

Bn геометрична прогресія, b1=-81 q=1/3 S4=?​

Answer 1

Объяснение:

b₁= -81       q=1/3

S₄=$\frac{-81*(1-(\frac{1}{3})^{4} ) }{1-\frac{1}{3} } =\frac{-81*(1-\frac{1}{81}) }{\frac{2}{3} } =\frac{-81*\frac{80}{81} }{\frac{2}{3} } =-80:\frac{2}{3} =-80*\frac{3}{2} = -120$

Answer 2

-120

Объяснение:

b₁= -81   q=1/3  S₄=?​

b₄=b₁ q³= -81 × (1/3)³= -81× 1/27= -3

$S_4=\frac{b_4q-b_1}{q-1} =\frac{-3*\frac{1}{3}-(-81) }{\frac{1}{3}-1 } =\frac{-1+81}{-\frac{2}{3} } = -80*\frac{3}{2} = -40*3= -120$