tui làm được câu c thui
c) (1-1/2).(1-1/3).(1-1/4).(1-1/5)...(1-1/2022).(1-1/2023)
= 1 2 3 4 2 3 4 5 . . . . . 2021 2022 2022 2023 = 1.2.3.4.5....2021.2022 2.3.4.5....2022.2023 = 1 2023

a: \(\frac{2022\cdot2023-2022}{2021\cdot2022+2022}\)

\(=\frac{2022\left(2023-1\right)}{2022\left(2021+1\right)}\)

\(=\frac{2022\cdot2022}{2022\cdot2022}=1\)

c: \(\left(1-\frac12\right)\left(1-\frac13\right)\cdot\ldots\cdot\left(1-\frac{1}{2022}\right)\left(1-\frac{1}{2023}\right)\)

\(=\frac12\cdot\frac23\cdot\ldots\cdot\frac{2021}{2022}\cdot\frac{2022}{2023}\)

\(=\frac{1}{2023}\)

Ta có: \(2023-\frac12\left(1+2\right)-\frac13\left(1+2+3\right)-\cdots-\frac{1}{2022}\left(1+2+\cdots+2022\right)\)

\(=2023-\frac12\cdot\frac{2\cdot3}{2}-\frac13\cdot\frac{3\cdot4}{2}-\cdots-\frac{1}{2022}\cdot\frac{2022\cdot2023}{2}\)

\(=2023-\frac32-\frac42-\cdots-\frac{2023}{2}=2023-\frac12\left(3+4+\cdots+2023\right)\)

\(=2023-\frac12\frac{\left(2023-3+1\right)\left(2023+3\right)}{2}=2023-\frac12\cdot\frac{2021\cdot2026}{2}=2023-\frac12\cdot2021\cdot1013\)

=-1021613,5

Ta có: C = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/2021.2022.2023

=> C = 1/2. (3-1/1.2.3 + 4-2/2.3.4 + 5-3/3.4.5 + ... + 2023-2021/2021.2022.2023

=> C = 1/2. (1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 + ... + 1/2021.2022 - 1/2022.2023)

=> C = 1/2. (1/1.2 - 1/2022.2023)

- Phần còn lại bạn tự tính chứ số to quá

B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + \(\dfrac{2022}{1}\)

B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + 2022

B = 1 + ( 1 + \(\dfrac{1}{2022}\)) + ( 1 + \(\dfrac{2}{2021}\)) + \(\left(1+\dfrac{3}{2020}\right)\)+ ... + \(\left(1+\dfrac{2021}{2}\right)\) 

B = \(\dfrac{2023}{2023}\) + \(\dfrac{2023}{2022}\) + \(\dfrac{2023}{2021}\) + \(\dfrac{2023}{2020}\) + ...+ \(\dfrac{2023}{2}\) 

B = 2023 \(\times\) ( \(\dfrac{1}{2023}\) + \(\dfrac{1}{2022}\) + \(\dfrac{1}{2021}\) + \(\dfrac{1}{2020}\)+ ... + \(\dfrac{1}{2}\))

Vậy B > C 

Ta có: \(B=\frac12+\frac13-\frac14+\frac15-\frac16+\cdots-\frac{1}{2022}+\frac{1}{2023}\)

=>\(B=\frac12+\frac13+\frac14+\frac15+\frac16+\cdots+\frac{1}{2022}+\frac{1}{2023}-2\left(\frac14+\frac16+\cdots+\frac{1}{2022}\right)\)

\(=\frac12+\frac13+\frac14+\frac15+\cdots+\frac{1}{2022}+\frac{1}{2023}-\frac12-\frac13-\cdots-\frac{1}{1011}\)

\(=\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2022}+\frac{1}{2023}\)

=C

=>B-C=0

Ta có; \(B=1-\frac12+\frac13-\frac14+\cdots-\frac{1}{2022}+\frac{1}{2023}\)

\(=1+\frac12+\frac13+\cdots+\frac{1}{2023}-2\left(\frac12+\frac14+\cdots+\frac{1}{2022}\right)\)

\(=1+\frac12+\ldots+\frac{1}{2023}-1-\frac12-\cdots-\frac{1}{1011}=\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2023}\)

=C

=>B-C=0

\(S=-\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+...+\dfrac{1}{5^{2022}}-\dfrac{1}{5^{2023}}\)

\(\Rightarrow\dfrac{25}{5}=-1+\dfrac{1}{5}-\dfrac{1}{5^2}+...+\dfrac{1}{5^{2021}}-\dfrac{1}{5^{2022}}\)

\(\Rightarrow5S+S=\left(-1+\dfrac{1}{5}-\dfrac{1}{5^2}+...+\dfrac{1}{5^{2021}}-\dfrac{1}{5^{2022}}\right)+\left(-\dfrac{1}{5}+\dfrac{1}{5^2}-...+\dfrac{1}{5^{2022}}-\dfrac{1}{5^{2023}}\right)\)

\(\Rightarrow6S=-1+\dfrac{1}{5}-\dfrac{1}{5^2}+...+\dfrac{1}{5^{2021}}-\dfrac{1}{5^{2022}}-\dfrac{1}{5}+\dfrac{1}{5^2}-...+\dfrac{1}{5^{2022}}-\dfrac{1}{5^{2023}}\)

\(\Rightarrow6S=-1-\dfrac{1}{5^{2023}}\)

\(\Rightarrow S=\dfrac{-1-\dfrac{1}{5^{2023}}}{6}\)

Ta có: \(\frac{2022}{1}+\frac{2021}{2}+\cdots+\frac{1}{2022}\)

\(=\left(\frac{2021}{2}+1\right)+\left(\frac{2020}{3}+1\right)+\cdots+\left(\frac{1}{2022}+1\right)+1\)

\(=\frac{2023}{2}+\frac{2023}{3}+\cdots+\frac{2023}{2023}=2023\left(\frac12+\frac13+\cdots+\frac{1}{2023}\right)\)

Ta có: \(\frac{\left(\frac12+\frac13+\cdots+\frac{1}{2023}\right)}{\frac{2022}{1}+\frac{2021}{2}+\cdots+\frac{1}{2022}}\)

\(=\frac{\left(\frac12+\frac13+\cdots+\frac{1}{2023}\right)}{2023\left(\frac12+\frac13+\cdots+\frac{1}{2023}\right)}\)

\(=\frac{1}{2023}\)

a:

Sửa đề: \(S=1-3+5-7+...+2021-2023+2025\)

Từ 1 đến 2025 sẽ có:

\(\dfrac{2025-1}{2}+1=\dfrac{2024}{2}+1=1013\left(số\right)\)

Ta có: 1-3=5-7=...=2021-2023=-2

=>Sẽ có \(\dfrac{1013-1}{2}=\dfrac{1012}{2}=506\) cặp có tổng là -2 trong dãy số này

=>\(S=506\cdot\left(-2\right)+2025=2025-1012=1013\)

b: \(S=1+2-3-4+5+6-7-8+...+2021+2022-2023-2024\)

Từ 1 đến 2024 là: \(\dfrac{\left(2024-1\right)}{1}+1=2024\left(số\right)\)

Ta có: 1+2-3-4=5+6-7-8=...=2021+2022-2023-2024=-4

=>Sẽ có \(\dfrac{2024}{4}=506\) cặp có tổng là -4 trong dãy số này

=>\(S=506\cdot\left(-4\right)=-2024\)