CMR:
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{\text{4}}+...+\frac{1}{2001}-\frac{1}{2002}=\frac{1}{1002}+...+\frac{1}{2002}\)
K
Khách
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28 tháng 10 2019
Câu hỏi của Cristiano Ronaldo - Toán lớp 7 - Học toán với OnlineMath
21 tháng 3 2018
\(1-\frac{1}{2}+\frac{1}{3}-...+\frac{1}{2001}-\frac{1}{2002}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2001}\right)\)\(-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2002}\right)\)
= \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}+\frac{1}{2002}\right)\)\(-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2002}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2002}\right)\)\(-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1001}\right)\)
\(=\frac{1}{1002}+\frac{1}{1003}+\frac{1}{1004}+...+\frac{1}{2002}\)
1 tháng 10 2016
Xét với n là số tự nhiên không nhỏ hơn 1
Ta có : \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Áp dụng điều trên ta có
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2002\sqrt{2001}+2001\sqrt{2002}}\)
\(=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2001}}-\frac{1}{\sqrt{2002}}\)
\(=1-\frac{1}{\sqrt{2002}}< 1-\frac{1}{\sqrt{2025}}=1-\frac{1}{45}=\frac{44}{45}\)
1 tháng 10 2016
ta chứng minh công thức tổng quát sau
\(\frac{1}{\left[n+1\right]\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left[n+1\right]}\left[\sqrt{n+1}+\sqrt{n}\right]}\)
=\(\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left[n+1\right]}\left[n+1-n\right]}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left[n+1\right]}}\)
=\(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
ta có \(\frac{1}{2\sqrt{1}+1\sqrt{2}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\)
\(\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)
........
\(\frac{1}{2002\sqrt{2001}+2001\sqrt{2002}}=\frac{1}{\sqrt{2001}}-\frac{1}{\sqrt{2002}}\)
=> \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+..+\frac{1}{2002\sqrt{2001}+2001\sqrt{2002}}\)
=\(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2001}}-\frac{1}{\sqrt{2002}}\)
=\(1-\frac{1}{\sqrt{2002}}< \frac{44}{45}\)
QD
26 tháng 1 2017
1)\(\frac{-8}{5}+\frac{207207}{201201}\)
=\(\frac{-8}{5}+\frac{207}{201}\)
=\(\frac{-8}{5}+\frac{69}{67}\)
=\(\frac{-191}{335}\)
3 tháng 8 2015
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2001}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2002}\right)\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2001}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2002}\right)-2\times\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2002}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{1001}+\frac{1}{1002}+...+\frac{1}{2001}+\frac{1}{2002}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1001}\right)\)
\(A=\frac{1}{1002}+\frac{1}{1003}+...\frac{1}{2002}\)= B
=> A/ B = 1
10 tháng 11 2018
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{2001^2}+\frac{1}{2002^2}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2000.2001}+\frac{1}{2001.2002}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{2000}-\frac{1}{2001}+\frac{1}{2001}-\frac{1}{2002}\)
\(\Rightarrow A< 1-\frac{1}{2002}=\frac{2001}{2002}\left(đpcm\right)\)
22 tháng 1 2016
\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2001}{2002}.\frac{2002}{2003}=\frac{1.2.3.....2001.2002}{2.3.4.....2002.2003}=\frac{1}{2003}\)
Ta có \(VT=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2001}-\frac{1}{2002}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2001}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2002}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}+\frac{1}{2002}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2002}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}+\frac{1}{2002}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1001}\right)\)
\(=\frac{1}{1002}+...\frac{1}{2002}=VP\)
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