(1/2-1).(1/3-1).(1/4-1).(1/5-1).....(1/2022-1).(1/2023-1).
híp me!
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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)
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
\(S=-\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+...+\dfrac{1}{5^{2022}}-\dfrac{1}{5^{2023}}\)
\(\Rightarrow\dfrac{25}{5}=-1+\dfrac{1}{5}-\dfrac{1}{5^2}+...+\dfrac{1}{5^{2021}}-\dfrac{1}{5^{2022}}\)
\(\Rightarrow5S+S=\left(-1+\dfrac{1}{5}-\dfrac{1}{5^2}+...+\dfrac{1}{5^{2021}}-\dfrac{1}{5^{2022}}\right)+\left(-\dfrac{1}{5}+\dfrac{1}{5^2}-...+\dfrac{1}{5^{2022}}-\dfrac{1}{5^{2023}}\right)\)
\(\Rightarrow6S=-1+\dfrac{1}{5}-\dfrac{1}{5^2}+...+\dfrac{1}{5^{2021}}-\dfrac{1}{5^{2022}}-\dfrac{1}{5}+\dfrac{1}{5^2}-...+\dfrac{1}{5^{2022}}-\dfrac{1}{5^{2023}}\)
\(\Rightarrow6S=-1-\dfrac{1}{5^{2023}}\)
\(\Rightarrow S=\dfrac{-1-\dfrac{1}{5^{2023}}}{6}\)
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
=>\(S=506\cdot\left(-4\right)=-2024\)