Chứng minh rằng \(17< \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}< 18\)
K
Khách
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Những câu hỏi liên quan
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 10 2019
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{\sqrt{n}}{n}+\frac{\sqrt{n+1}}{n+1}\)
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=\frac{\sqrt{1}}{1}-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{3}+...+\frac{\sqrt{99}}{99}-\frac{\sqrt{100}}{100}\)
\(=1-\frac{\sqrt{100}}{100}=\frac{9}{10}< 1\)
15 tháng 6 2018
Ta có:
\(\frac{1}{n\sqrt{\left(n+1\right)}+\left(n+1\right)\sqrt{n}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n}+\sqrt{\left(n+1\right)}\right)}\)
\(=\frac{1}{\sqrt{n\left(n+1\right)}}.\left(\sqrt{n+1}-\sqrt{n}\right)=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thế vào ta được
\(\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+...+\frac{1}{99\sqrt{100}+100\sqrt{99}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)
\(=1-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)
Y
16 tháng 6 2019
+ \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)-n}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Do đó : \(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)
\(=1-\frac{1}{10}=\frac{9}{10}\)
KN
18 tháng 10 2020
Ta có công thức tổng quát: \(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}\)(*)
Áp dụng (*), ta được: \(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}=\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{4}-\sqrt{3}\right)+...+\left(\sqrt{100}-\sqrt{99}\right)=\sqrt{100}-\sqrt{1}=9\left(đpcm\right)\)
18 tháng 10 2020
Trục căn thức ở mẫu :
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\frac{\sqrt{1}-\sqrt{2}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{1}-\sqrt{2}\right)}+\frac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}+\frac{\sqrt{3}-\sqrt{4}}{\left(\sqrt{3}+\sqrt{4}\right)\left(\sqrt{3}-\sqrt{4}\right)}+...+\frac{\sqrt{99}-\sqrt{100}}{\left(\sqrt{99}+\sqrt{100}\right)\left(\sqrt{99}-\sqrt{100}\right)}\)
\(=\frac{\sqrt{1}-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+\frac{\sqrt{3}-\sqrt{4}}{3-4}+...+\frac{\sqrt{99}-\sqrt{100}}{99-100}\)
\(=\frac{\sqrt{1}-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+\frac{\sqrt{3}-\sqrt{4}}{-1}+...+\frac{\sqrt{99}-\sqrt{100}}{-1}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{100}-\sqrt{99}\)
\(=\sqrt{100}-\sqrt{1}\)
\(=10-1=9\)
=> đpcm
+ \(2\cdot\frac{1}{\sqrt{n}+\sqrt{n+1}}< \frac{2}{\sqrt{n}+\sqrt{n}}< 2\cdot\frac{1}{\sqrt{n-1}+\sqrt{n}}\) \(\Rightarrow2\left(\sqrt{n+1}-\sqrt{n}\right)< \frac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)
\(\Rightarrow A>2\left(\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{101}-\sqrt{100}\right)\)
\(\Rightarrow A>2\left(\sqrt{101}-\sqrt{2}\right)>17\)
+ \(A< 2\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\right)\Rightarrow A< 2\left(\sqrt{100}-1\right)=18\)