Tính `1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+…+1/(1+2+3+4+…+2021)`.

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1. Đáp án: `≈ 2`

   Giải thích các bước giải:

   `1+“\frac{1}{1+2}“+“\frac{1}{1+2+3}“+…+“\frac{1}{1+2+3+…+2021}`

   `=1+“\frac{1}{2.3:2}“+“\frac{1}{3.4:2}“+“\frac{1}{4.5:2}“+…+“\frac{1}{2021.2022:2}`

   `=1+“(“\frac{2}{2.3}“+“\frac{2}{3.4}“+“\frac{2}{4.5}“+…+“\frac{2}{2021.2022}“)`

   `=1+2(1/2-1/3+1/3-1/4+1/4-1/5+…+“\frac{1}{2021}“-“\frac{1}{2022}“)`

   `=1+2(1/2-1/2022)`

   `=1+2.“\frac{505}{1011}`

   `=1+1010/1011` `≈ 2`

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2. Đặt `A = 1 + 1/(1 + 2) + 1/(1 + 2 + 3) + 1/(1 + 2 + 3 + 4) + … + 1/(1 + 2 + 3 + 4 + … + 2021)`

   Ta có:

   `1/(1 + 2) = 1/3 = 2/(2. 3)`

   `1/(1 + 2 + 3) = 1/6 = 2/12 = 2/(3. 4)`

   `1/(1 + 2 +3 + 4) = 1/10 = 2/20 = 2/(4. 5)`

   `…`

   `1/(1 + 2 + 3 + 4 + … + 2021) = 1/2043231 = 2/4086462 = 2/(2021. 2022)`

   `=> A = 1 + 2/(2. 3) + 2/(3. 4) + 2/(4. 5) + … + 2/(2021. 2022)`

   `=> A = 1 + 2. (1/(2. 3) + 1/(3. 4) + 1/(4. 5) + … + 1/(2021. 2022))`

   `=> A = 1 + 2. (1/2 – 1/3 + 1/3 – 1/4 + 1/4 – 1/5 + … + 1/2021 – 1/2022)`

   `=> A + 1 + 2. (1/2 – 1/2022)`

   `=> A = 1 + 1 – 1/1011`

   `=> A = 2 – 1/1011`

   `=> A = 2022/1011 – 1/1011`

   `=> A = 2021/1011`

   Vậy `A = 2021/1011`

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