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

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

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

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

\(\Rightarrow\frac{A}{B}=\frac{1}{n}\).

chỗ \(\sqrt{n}-\sqrt{n+1}\)phải là \(\sqrt{n}+\sqrt{n+1}\)

a, Ta có

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

\(=\frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{2n+1}=\frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{4n^2+4n+1}}< \frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{4n^2+4n}}\)

mà \(\frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{4n^2+4n}}=\frac{2\cdot\left(\sqrt{n+1}-\sqrt{n}\right)}{2\sqrt{n\left(n+1\right)}}=\frac{\sqrt{n+1}}{\sqrt{n}\cdot\sqrt{n+1}}-\frac{\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}}\)

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

b, áp dụng bđt ta có

\(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{4023\cdot\left(\sqrt{2011}+\sqrt{2012}\right)}< \frac{2011}{2013}\)

\(=\frac{1}{\left(2\cdot1+1\right)\left(1+\sqrt{2}\right)}+\frac{1}{\left(2\cdot2+1\right)\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{\left(2\cdot2011+1\right)\left(\sqrt{2011}-\sqrt{2012}\right)}\)

\(< 1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}\)..

\(=1-\frac{1}{\sqrt{2012}}=\frac{\sqrt{2012}-1}{\sqrt{2012}}=\frac{2011}{\sqrt{2012}\cdot\left(\sqrt{2012}+1\right)}\)

\(=\frac{2011}{2012+\sqrt{2012}}< \frac{2011}{2013}\)

Bạn Nhật sai đề bài

Câu. a. Dòng thứ nhất xuống dòng thứ 2. Em chú ý mẫu số sai rồi.

b. Công thức có số 2 trên tử số. Mà em ko đưa số 2 vào thì sao áp dụng dc công thức?