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Dana2554326
@Dana2554326
November 2021
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(1/3)^log3(1+1/4+1/16...)
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Dимасuk
Verified answer
1 + 1/4 + 1/16 + ... - это сумма бесконечно убывающей геометрической прогрессии
b₁ = 1; q = 1/4
S = b₁/(1 - q) = 1/(1 - 1/4) = 1/(3/4) = 4/3
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Dana2554326
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Answers & Comments
Verified answer
1 + 1/4 + 1/16 + ... - это сумма бесконечно убывающей геометрической прогрессииb₁ = 1; q = 1/4
S = b₁/(1 - q) = 1/(1 - 1/4) = 1/(3/4) = 4/3