Question

[Huy Tống Văn]

Chứng minh: a) 1/căn 1+1/căn 2+1/căn 3+…………+1/căn 100>18<19

Answer 1

[Nguyễn Việt Lâm]

Ý bạn là \(18< \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}< 19\) ?

Ta có:

\(A=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}=\frac{2}{2\sqrt{1}}+\frac{2}{2\sqrt{2}}+...+\frac{2}{2\sqrt{100}}\)

\(\Rightarrow A>\frac{2}{1+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{100}+\sqrt{101}}\)

\(\Rightarrow A>\frac{2\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\frac{2\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\frac{2\left(\sqrt{101}-\sqrt{100}\right)}{\left(\sqrt{101}-\sqrt{100}\right)\left(\sqrt{101}+\sqrt{100}\right)}\)

\(\Rightarrow A>2\left(\sqrt{2}-1+\sqrt{3}-2+...+\sqrt{101}-\sqrt{100}\right)\)

\(\Rightarrow A>2\left(\sqrt{101}-1\right)>2\left(\sqrt{100}-1\right)=18\)

Tương tự:

\(A=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}=1+\frac{2}{2\sqrt{2}}+\frac{2}{2\sqrt{3}}+...+\frac{2}{2\sqrt{100}}\)

\(\Rightarrow A< 1+\frac{2}{1+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{99}+\sqrt{100}}\)

Nhân liên hợp tử mẫu và rút gọn ta được (giống chứng minh >18 bên trên):

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

\(\Rightarrow A< 1+2\left(\sqrt{100}-1\right)=1+18=19\)

\(\Rightarrow18< A< 19\) (đpcm)

Answer 2

[Huy Tống Văn]

giup minh vs