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

Question

cattuong · Answer

Đ&aacute;p &aacute;n: `&asymp; 2`   Giải th&iacute;ch c&aacute;c 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` `&asymp; 2`

maianhnhi · Answer

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`

Trả lời

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

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