Ta có : \(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)

\(=\frac{\sqrt{1}-\sqrt{2}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{1}-\sqrt{2}\right)}+\frac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}+...+\frac{\sqrt{n-1}-\sqrt{n}}{\left(\sqrt{n-1}+\sqrt{n}\right)\left(\sqrt{n-1}-\sqrt{n}\right)}\)

\(=\frac{\sqrt{1}-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+...+\frac{\sqrt{n-1}-\sqrt{n}}{n-1-n}\)

\(=\frac{\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+...+\sqrt{n-1}-\sqrt{n}}{-1}\)

\(=\frac{\sqrt{1}-\sqrt{n}}{-1}=\sqrt{n}-\sqrt{1}=\sqrt{n}-1\)

a)\(\frac{1}{\left(n+1\right).\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2.n-n^2\left(n+1\right)}\)

\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

b)\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)

\( S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}\)

\(a,\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}\cdot\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)}\)

\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)

\(=\frac{\sqrt{n-1}-\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}\left(n+1-n\right)}\)

\(=\frac{\sqrt{n-1}-\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}}=\frac{\sqrt{n-1}}{\sqrt{n}\cdot\sqrt{n+1}}-\frac{\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}}\)

\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

b, \(S=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{3\sqrt{4}+4\sqrt{3}}+...+\frac{1}{400\sqrt{399}+399\sqrt{400}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}\)

Đề sai '-'

 Xét số hạng tổng quát ta có:

\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{\left(n+1\right)n}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)

\(=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)< \sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

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

Áp dụng vào bài tập, ta có:

\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}\)

\(< \frac{2}{\sqrt{1}}-\frac{2}{\sqrt{2}}+\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{3}}+...+\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\)

\(=2-\frac{2}{\sqrt{n+1}}< 2\left(đpcm\right)\)

Đặt:

\(A=\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}\)

\(\Leftrightarrow2A=\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}+\frac{1}{\sqrt{97}+\sqrt{99}}\)

\(>\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}+\frac{1}{\sqrt{99}+\sqrt{101}}\)

\(=\frac{1}{2}.\left(\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+...+\sqrt{101}-\sqrt{99}\right)\)

\(=\frac{1}{2}.\left(\sqrt{101}-\sqrt{1}\right)>\frac{1}{2}.\left(\sqrt{100}-\sqrt{1}\right)\)

\(=\frac{9}{2}\)

\(\Rightarrow A>\frac{9}{4}\)

Câu 2/ Ta có:

\(n^{n+1}>\left(n+1\right)^n\)

\(\Leftrightarrow n>\left(1+\frac{1}{n}\right)^n\left(1\right)\)

Giờ ta chứng minh cái (1) đúng với mọi \(n\ge3\)

Với \(n=3\) thì dễ thấy (1) đúng.

Giả sử (1) đúng đến \(n=k\) hay

\(k>\left(1+\frac{1}{k}\right)^k\)

Ta cần chứng minh (1) đúng với \(n=k+1\)hay \(k+1>\left(1+\frac{1}{k+1}\right)^{k+1}\)

Ta có: \(\left(1+\frac{1}{k+1}\right)^{k+1}< \left(1+\frac{1}{k}\right)^{k+1}=\left(1+\frac{1}{k}\right)^k.\left(1+\frac{1}{k}\right)\)

\(< k\left(1+\frac{1}{k}\right)=k+1\)

Vậy có ĐPCM

bằng 122223

Lời giải:

Xét số hạng tổng quát:

\(\frac{1}{(n+1)\sqrt{n+1}+n\sqrt{n}}=\frac{1}{(\sqrt{n}+\sqrt{n+1})[n+\sqrt{n(n+1)}+n+1)]}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{(\sqrt{n}+\sqrt{n+1})[n+\sqrt{n(n+1)}+n+1)]}\)

\(=\frac{\sqrt{n+1}-\sqrt{n}}{n+(n+1)-\sqrt{n(n+1)}}<\frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n(n+1)}-\sqrt{n(n+1)}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Do đó:

\(\frac{1}{2\sqrt{2}+1\sqrt{1}}< \frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\)

\(\frac{1}{3\sqrt{3}+2\sqrt{2}}< \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)

......

\(\frac{1}{(n+1)\sqrt{n+1}+n\sqrt{n}}< \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Cộng theo vế:

\(\Rightarrow \text{VT}< 1-\frac{1}{\sqrt{n+1}}\) (đpcm)