Question

[༺ ๖ۣۜPhạm ✌Tuấn ✌Kiệτ ༻]

Chứng minh rằng \(17< \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}< 18\)

Answer 1

[Sách Giáo Khoa]

Answer 2

[bach nhac lam]

+ \(2\cdot\frac{1}{\sqrt{n}+\sqrt{n+1}}< \frac{2}{\sqrt{n}+\sqrt{n}}< 2\cdot\frac{1}{\sqrt{n-1}+\sqrt{n}}\) \(\Rightarrow2\left(\sqrt{n+1}-\sqrt{n}\right)< \frac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)

\(\Rightarrow A>2\left(\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{101}-\sqrt{100}\right)\)

\(\Rightarrow A>2\left(\sqrt{101}-\sqrt{2}\right)>17\)

+ \(A< 2\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\right)\Rightarrow A< 2\left(\sqrt{100}-1\right)=18\)