Trang chủ » Hỏi đáp » Môn Toán tính `1 – 3/4 + (3/4)^2 – (3/4)^3 + (3/4)^4 – … – (3/4)^2019 + (3/4)^2020` 03/03/2024 tính `1 – 3/4 + (3/4)^2 – (3/4)^3 + (3/4)^4 – … – (3/4)^2019 + (3/4)^2020`
Giải đáp + Lời giải và giải thích chi tiết: Đặt A=1 – 3/4 + (3/4)^2 – (3/4)^3 + (3/4)^4 – … – (3/4)^2019 + (3/4)^2020 =>3/4A=3/4- (3/4)^2 + (3/4)^3 – (3/4)^4 + (3/4)^5 – … – (3/4)^2020 + (3/4)^2021 =>3/4A+A=[3/4- (3/4)^2 + (3/4)^3 – (3/4)^4 + (3/4)^5 – … – (3/4)^2020 + (3/4)^2021]+[1 – 3/4 + (3/4)^2 – (3/4)^3 + (3/4)^4 – … – (3/4)^2019 + (3/4)^2020] =>7/4A=(3/4)^2021+1 =>A=4/7[(3/4)^2021+1] Vậy 1 – 3/4 + (3/4)^2 – (3/4)^3 + (3/4)^4 – … – (3/4)^2019 + (3/4)^2020=4/7[(3/4)^2021+1] Trả lời
Giải đáp + Lời giải và giải thích chi tiết: Đặt A = 1 – 3/4 + (3/4)^2 – (3/4)^3 + (3/4)^4 – … – (3/4)^2019 + (3/4)^2020 3/4A = 3/4 – (3/4)^2 + (3/4)^3 – (3/4)^4 – … – (3/4)^2020 + (3/4)^2021 A + 3/4A = (1 – 3/4 + (3/4)^2 – (3/4)^3 + (3/4)^4 – … – (3/4)^2019 + (3/4)^2020) + (3/4 – (3/4)^2 + (3/4)^3 – (3/4)^4 – … – (3/4)^2020 + (3/4)^2021) 7/4A = 1 – 3/4 + (3/4)^2 – (3/4)^3 + (3/4)^4 – … – (3/4)^2019 + (3/4)^2020 + 3/4 – (3/4)^2 + (3/4)^3 – (3/4)^4 – … – (3/4)^2020 + (3/4)^2021 7/4A = 1 – (3/4)^2021 A = (1 – (3/4)^2021)/(7/4). Trả lời
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