Trang chủ » Hỏi đáp » Môn Toán 1/2+1/2^2+1/2^3+……+1/2^2021+1/2^2022 07/07/2023 1/2+1/2^2+1/2^3+……+1/2^2021+1/2^2022
Đặt A=1/2+1/2^2+1/2^3+…+1/2^2021+1/2^2022 =>2A=1+1/2+1/2^2+…+1/2^2020+1/2^2021 =>2A-A=(1+1/2+1/2^2+…+1/2^2020+1/2^2021)-(1/2+1/2^2+1/2^3+……+1/2^2021+1/2^2022) =>A=1+1/2+1/2^2+…+1/2^2020+1/2^2021-1/2-1/2^2-1/2^3+……-1/2^2021-1/2^2022 =>A=1-1/2^2022 Vậy A=1-1/2^2022 (Nếu đề yêu cầu rút gọn, tus không nói kĩ nha :))) Trả lời
Đặt: S=1/2+ 1/2^2+ 1/2^3+…+ 1/2^2021+ 1/2^2022 ⇒2S= 1+ 1/2+ 1/2^2+…+ 1/2^2020+ 1/2^2021 ⇒2S-S= (1+ 1/2+…+ 1/2^2021)-(1/2+ 1/2^2+…+ 1/2^2022) ⇒ S= 1- 1/2^2022 ⇒ S= (2^2022- 1)/2^2022 Vậy, S= (2^2022- 1)/2^2022 ~ $kiddd$ ~ Trả lời
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