Đáp án:
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Đề ` 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^{2020}`
Đặt `A = 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^{2020}`
`↔ 3A = 3 (1/3 + 1/3^2 + 1/3^3 + ... + 1/3^{2020})`
`↔ 3A = 1 + 1/3 + 1/3^2 + ... + 1/3^{2019}`
`↔ 3A - A = (1 + 1/3 + 1/3^2 + ... + 1/3^{2019})- (1/3 + 1/3^2 + 1/3^3 + ... + 1/3^{2020})`
`↔ (3-1) A= 1 - 1/3^{2020}`
`↔ 2A =1 - 1/3^{2020}`
`↔ A = (1 - 1/3^{2020})/2`
Vậy `A = (1 - 1/3^{2020})/2`
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Đề `1/2 + 1/2^2 + 1/2^3 + ... + 1/2^{2020}`
Đặt `A = 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^{2020}`
`↔ 2A = 2 (1/2 + 1/2^2 + 1/2^3 + ... + 1/2^{2020})`
`↔ 2A = 1 + 1/2 + 1/2^2 + ... + 1/2^{2019}`
`↔ 2A - A = (1 + 1/2 + 1/2^2 + ... + 1/2^{2019}) - (1/2 + 1/2^2 + 1/2^3 + ... + 1/2^{2020})`
`↔ (2-1)A = 1 - 1/2^{2020}`
`↔ A = 1 - 1/2^{2020}`
Vậy `A = 1- 1/2^{2020}`
