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BT
0
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 1
\(A< \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{2003.2004}\)
\(\Rightarrow A< \dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{2003}-\dfrac{1}{2004}\)
\(\Rightarrow A< \dfrac{1}{4}-\dfrac{1}{2004}< \dfrac{1}{4}\)
Đồng thời:
\(A>\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{2004.2005}\)
\(A>\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{2004}-\dfrac{1}{2005}\)
\(A>\dfrac{1}{5}-\dfrac{1}{2005}=\dfrac{80}{401}>\dfrac{50}{500}>\dfrac{1}{10}>\dfrac{1}{65}\)
Vậy \(\dfrac{1}{65}< A< \dfrac{1}{4}\)
TB
1
TM
15 tháng 3 2017
Bài toán tổng quát:
Với mọi n\(\in\)N* ta có: \(\frac{1}{n^3}< \frac{1}{n^3-n}=\frac{1}{n\left(n^2-1\right)}=\frac{1}{\left(n-1\right)n\left(n+1\right)}\)
Áp dụng vào bài toán:
\(\frac{1}{5^3}+\frac{1}{6^3}+\frac{1}{7^3}+...+\frac{1}{2004^3}< \frac{1}{4.5.6}+\frac{1}{5.6.7}+\frac{1}{6.7.8}+...+\frac{1}{2003.2004.2005}\)
mà \(\frac{1}{4.5.6}+\frac{1}{5.6.7}+\frac{1}{6.7.8}+...+\frac{1}{2003.2004.2005}\)
\(=\frac{1}{2}\left(\frac{2}{4.5.6}+\frac{2}{5.6.7}+\frac{2}{6.7.8}...+\frac{2}{2003.2004.2005}\right)\)
\(=\frac{1}{2}\left(\frac{1}{4.5}-\frac{1}{5.6}+\frac{1}{5.6}-\frac{1}{6.7}+\frac{1}{6.7}-\frac{1}{7.8}...+\frac{1}{2003.2004}-\frac{1}{2004.2005}\right)\)
\(=\frac{1}{2}\left(\frac{1}{4.5}-\frac{1}{2003.2004}\right)=\frac{1}{40}-\frac{1}{2.2003.2004}< \frac{1}{40}\)
=>\(\frac{1}{3.4.5}+\frac{1}{4.5.6}+\frac{1}{5.6.7}+...+\frac{1}{2002.2003.2004}< \frac{1}{40}\)
TX
1
TB
0
29 tháng 3 2023
A=1/5^3+1/6^3+...+1/2023^3
1/5^3<1/4*5*6
Xét tương tự, ta đều sẽ được:
\(\dfrac{1}{n^3}< \dfrac{1}{n\left(n-1\right)\left(n+1\right)}\)
=>\(A< \dfrac{1}{4\cdot5\cdot6}+\dfrac{1}{5\cdot6\cdot7}+...+\dfrac{1}{2022\cdot2023\cdot2024}\)
=>\(A< \dfrac{1}{2}\left(\dfrac{2}{4\cdot5\cdot6}+\dfrac{2}{5\cdot6\cdot7}+...+\dfrac{2}{2022\cdot2023\cdot2024}\right)\)
=>\(A< \dfrac{1}{2}\left(\dfrac{1}{4\cdot5}-\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}-\dfrac{1}{7\cdot8}+...+\dfrac{1}{2022\cdot2023}-\dfrac{1}{2023\cdot2024}\right)\)
=>A<1/40
Ta có BĐT: \(\dfrac{1}{k\left(k+1\right)\left(k+2\right)}< \dfrac{1}{k^3}< \dfrac{1}{\left(k-1\right)\cdot k\cdot\left(k+1\right)}\)
Do đó, ta được:
\(\dfrac{1}{5\cdot6\cdot7}+\dfrac{1}{6\cdot7\cdot8}+...+\dfrac{1}{2023\cdot2024\cdot2025}< A\)
\(\Leftrightarrow A>\dfrac{1}{2}\left(\dfrac{1}{5\cdot6}-\dfrac{1}{2024\cdot2025}\right)>\dfrac{1}{2}\left(\dfrac{1}{30}-\dfrac{1}{390}\right)=\dfrac{1}{65}\)
=>1/65<A<1/40
12 tháng 2 2018
Ta có: \(n^3-n< n^3\forall n\)
mà: \(n^3-n=n\left(n^2-1\right)=\left(n-1\right)n\left(n+1\right)\)
Nên: \(\left(n-1\right)n\left(n+1\right)< n^3\Leftrightarrow\dfrac{1}{\left(n-1\right)n\left(n+1\right)}>\dfrac{1}{n^3}\)
Trở lại bài toán:
\(SV=\dfrac{1}{5^3}+\dfrac{1}{6^3}+\dfrac{1}{7^3}+...+\dfrac{1}{2004^3}< \dfrac{1}{\left(5-1\right).5.\left(5+1\right)}+\dfrac{1}{\left(6-1\right).6.\left(6+1\right)}+\dfrac{1}{\left(7-1\right).7.\left(7+1\right)}+...+\dfrac{1}{\left(2004-1\right).2004.\left(2004+1\right)}\)
\(SV< \dfrac{1}{4.5.6}+\dfrac{1}{5.6.7}+\dfrac{1}{6.7.8}+...+\dfrac{1}{2003.2004.2005}=\dfrac{1}{2}\left(\dfrac{1}{4.5}-\dfrac{1}{5.6}+\dfrac{1}{5.6}-\dfrac{1}{6.7}+\dfrac{1}{6.7}-\dfrac{1}{7.8}+...+\dfrac{1}{2003.2004}-\dfrac{1}{2004.2005}\right)=\dfrac{1}{2}\left(\dfrac{1}{4.5}-\dfrac{1}{2004.2005}\right)=\dfrac{1}{2.4.5}-\dfrac{1}{2.2004.2005}=\dfrac{1}{40}-\dfrac{1}{2.2004.2005}< \dfrac{1}{40}\left(đpcm\right)\)