c=1-1/2 +1/3+1/4+...+1/2023+1/2024 phần 1/1013+1/1014+1/1015+...+1/2025
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NL
Nguyễn Lê Phước Thịnh
CTVHS
11 tháng 11 2025
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
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
1 tháng 11 2024
A = \(\dfrac{1}{2021.2022}\) + \(\dfrac{1}{2022.2023}\) + \(\dfrac{1}{2023.2024}\) + \(\dfrac{1}{2024.2025}\) - \(\dfrac{4}{2021.2025}\)
A = \(\dfrac{1}{2021}\) - \(\dfrac{1}{2022}\) + \(\dfrac{1}{2022}\) - \(\dfrac{1}{2023}\) + \(\dfrac{1}{2023}\) - \(\dfrac{1}{2024}\) + \(\dfrac{1}{2024}\) - \(\dfrac{1}{2025}\) - \(\dfrac{1}{2021}\) + \(\dfrac{1}{2025}\)
A = (\(\dfrac{1}{2021}\) - \(\dfrac{1}{2021}\)) + (\(\dfrac{1}{2022}\) - \(\dfrac{1}{2022}\)) + (\(\dfrac{1}{2023}\) - \(\dfrac{1}{2023}\)) + (\(\dfrac{1}{2024}\) - \(\dfrac{1}{2024}\)) + (\(\dfrac{1}{2025}\) - \(\dfrac{1}{2025}\))
A = 0 + 0 +0 + 0+ ... + 0
A = 0
10 tháng 9 2023
\(S=C^0_{2024}+\dfrac{1}{2}C^2_{2024}+\dfrac{1}{3}C^4_{2024}+\dfrac{1}{4}C^6_{2024}+...+\dfrac{1}{1013}C^{2024}_{2024}\)
Ta có :
\(\dfrac{1}{k+1}C^{2k-1}_n=\dfrac{1}{k+1}.\dfrac{n!}{\left(2k-1\right)!\left(n-2k+1\right)!}\)
\(=\dfrac{1}{n+1}.\dfrac{\left(n+1\right)!}{2k!\left[\left(n+1\right)-2k\right]!}\)
\(=\dfrac{1}{n+1}C^{2k}_{n+1}\)
\(\Rightarrow S_n=\dfrac{1}{n+1}\Sigma^{2k}_{k=0}C^{2k}_{n+1}=\dfrac{1}{n+1}\left(\Sigma^{2k}_{k=0}C^{2k-1}_{n+1}-C^0_{n+1}\right)=\dfrac{2^{2n-1}-1}{n+1}\)
\(\Rightarrow S=\dfrac{2^{2025}-1}{1013}\)
M
10 tháng 9 2023
S = C₀₂₀₂₄ + 12.C₂₀₂₄ + 13.C₂₀₂₄ + 14.C₂₀₂₄ + ... + 11013.C₂₀₂₄
= (C₀₂₀₂₄ + C₂₀₂₄ + C₂₀₂₄ + C₂₀₂₄ + ... + C₂₀₂₄) + (C₂₀₂₄ + C₂₀₂₄ + C₂₀₂₄ + ... + C₂₀₂₄) + ... + (C₂₀₂₄)
= 11014.C₂₀₂₄
= 11014.
1 tháng 5 2017
\(\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...\left(1-\frac{1015}{1014}\right)\)
\(=\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...\left(1-\frac{1014}{1014}\right).\left(1-\frac{1015}{1014}\right)\)
\(=\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...\left(1-1\right).\left(1-\frac{1015}{1014}\right)\)
\(=\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...0.\left(1-\frac{1015}{1014}\right)\)
\(=0\)
Ta có
\(c = 1 - \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \hdots + \frac{1}{2023} + \frac{1}{2024}\)
Gọi
\(S = 1 + \frac{1}{2} + \frac{1}{3} + \hdots + \frac{1}{2024}\)
thì
\(c = S - 2 \cdot \frac{1}{2} = S - 1\)
Vì \(S\) là tổng điều hoà:
\(S = H_{2024}\)
với \(H_{n} \approx ln n + \gamma\)
Trong đó \(\gamma \approx 0.577\)
Tính gần đúng
\(H_{2024} \approx ln \left(\right. 2024 \left.\right) + 0.577\) \(ln \left(\right. 2024 \left.\right) \approx 7.61\) \(H_{2024} \approx 7.61 + 0.577 = 8.187\)
Do đó
\(c \approx 8.187 - 1 = 7.187\)
Phần cần tính thêm
\(\frac{1}{1013} + \frac{1}{1014} + \hdots + \frac{1}{2025}\)
Gọi:
\(T = H_{2025} - H_{1012}\)
Xấp xỉ:
\(H_{n} \approx ln n + 0.577\) \(T \approx ln \left(\right. 2025 \left.\right) - ln \left(\right. 1012 \left.\right)\) \(= ln \textrm{ } \left(\right. \frac{2025}{1012} \left.\right)\) \(\frac{2025}{1012} \approx 2\) \(T \approx ln 2 \approx 0.693\)
✅ Kết quả gần đúng:
\(c \approx 7.19\) \(\frac{1}{1013} + \hdots + \frac{1}{2025} \approx 0.693\)