ghi rõ đề bài ra nhanh lên

Ta có: \(2y+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{2020\cdot2021}\right)=\dfrac{4041}{2021}\)

\(\Leftrightarrow2y+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2020}-\dfrac{1}{2021}\right)=\dfrac{4041}{2021}\)

\(\Leftrightarrow2y+1-\dfrac{1}{2021}=\dfrac{4041}{2021}\)

\(\Leftrightarrow2y=\dfrac{4041}{2021}+\dfrac{1}{2021}-1\)

\(\Leftrightarrow2y=2-1=1\)

hay \(y=\dfrac{1}{2}\)

nhanh nha các bạn

Nguồn: Tính tổng: 1x2 + 2x3 + 3x4 +...+ 2019x2020 + 2020x2021 - Hoc24
Đặt $A=1.2+2.3+3.4+.........+2019.2020+2020.2021$

$\Rightarrow 3A=\mathrm{1.2.3}+\mathrm{2.3.3}+\mathrm{3.4.3}+.....+\mathrm{2019.2020.3}+\mathrm{2020.2021.3}$

$=\mathrm{1.2.3}+\mathrm{2.3.}\left(4-1\right)+\mathrm{3.4.}\left(5-2\right)+.....+\mathrm{2020.2021.}\left(2022-2019\right)$

$=\mathrm{1.2.3}+\mathrm{2.3.4}-\mathrm{1.2.3}+\mathrm{3.4.5}-\mathrm{2.3.4}+...+\mathrm{2020.2021.2022}-\mathrm{2019.2020.2021}$

$=\mathrm{2020.2021.2022}$

$\Rightarrow A=\frac{}{}$

cho mik hoi ket qua la bao nhieu

Đặt A = 1.2 + 2.3 + 3.4 + ... + 2019.2020 + 2020.2021

=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 2019.2020.3 + 2020.2021.3

=> 3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2019.2020.(2021 - 2018) + 2020.2021.(2022 - 2019)

=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 2019.2020.2021 - 2018.2019.2020 + 2020.2021.2022 - 2019.2020.2021

=> 3A = 2020.2021.2022

=> A = 2 751 551 080

Đặt \(A=1.2+2.3+3.4+.........+2019.2020+2020.2021\)

\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+.....+2019.2020.3+2020.2021.3\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+.....+2020.2021.\left(2022-2019\right)\)

\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+2020.2021.2022-2019.2020.2021\)

\(=2020.2021.2022\)

\(\Rightarrow A=\frac{2020.2021.2022}{3}\)

ai nhanh mik tik cho please ☹☹ω

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Ta có: \(A=\frac{1\times2}{2\times2}\times\frac{2\times3}{3\times3}\times\frac{3\times4}{4\times4}\times\ldots\times\frac{2019\times2020}{2020\times2020}\)

\(=\frac12\times\frac23\times\frac34\times\ldots\times\frac{2019}{2020}=\frac{1}{2020}\)

ta có: \(B=\frac{2020\times2021-2021\times2019}{2021\times2017+2021\times3}\)

\(=\frac{2021\times\left(2020-2019\right)}{2021\times\left(2017+3\right)}=\frac{1}{2020}\)

Do đó: A=B