Знайти суму перших п'яти членів геометричної прогресії (bn), у якій b4=1/16, b5=1/64. Created by Олесяя1111111. algebra-ru

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Амонита $b _{4}= \frac{1}{16}\\b _{5} = \frac{1}{64} \\\\ b_{5} = b_{4} *q\\\\q= \frac{ b_{5} }{ b_{4} }= \frac{ \frac{1}{64} }{ \frac{1}{16} }= \frac{1}{4}\\\\ b_{4} = b_{1}* q^{3} \\\\ b_{1}= \frac{ b_{4} }{ q^{3} }= \frac{ \frac{1}{16} }{( \frac{1}{4}) ^{3} } = \frac{ \frac{1}{16} }{ \frac{1}{64} }=4\\\\S _{5}= \frac{ b_{5} q- b_{1} }{q-1}= \frac{ \frac{1}{64}* \frac{1}{4}-4 }{ \frac{1}{4}-1 } = \frac{ \frac{1}{256}-4 }{- \frac{3}{4} } =- \frac{1023*4}{256*3} =-5 \frac{21}{64}$

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B5 = b4 · q → q = 1/64 : 1/16 = 1/4 bn =b1 · q^(n -1)n = 5b5 = b1 · q^4 → b1 = 1/64 : (1/4)^4 = 1/64 · 256 = 4 Sn = b1 · (q^n - 1) / (q - 1)n = 5S5 = (b5 · q - b1) / (q - 1)S5 = (1/64 · 1/4 - 4) / (1/4 - 1) = 341/64 = 5 21/64 = 5,328125Ответ: 5 21/64 или 5,328125

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