Chắc có lẽ bạn định làm như này:

\(\frac{1}{\left(3n+2\right)\left(3n+5\right)}=\frac{3}{3\left(3n+2\right)\left(3n+5\right)}=\frac{\left(3n+5\right)-\left(3n+2\right)}{3\left(3n+2\right)\left(3n+5\right)}=\frac{1}{3}\left[\frac{1}{3n+2}-\frac{1}{3n+5}\right]\)

\(A=\left(1+\frac{2}{4}\right)\left(1+\frac{2}{10}\right)...\left(1+\frac{2}{n^2+3n}\right)\)

\(A=\left(\frac{6}{4}\right)\left(\frac{12}{10}\right)...\left(\frac{n^2+3n+2}{n^2+3n}\right)\)

\(A=\left(\frac{2.3}{1.4}\right)\left(\frac{3.4}{2.5}\right)\left(\frac{4.5}{3.6}\right)...\left(\frac{\left(n+1\right)\left(n+2\right)}{n\left(n+3\right)}\right)\)

\(A=\frac{2.3.4...\left(n+1\right)}{1.2.3...n}.\frac{3.4.5...\left(n+2\right)}{4.5.6...\left(n+3\right)}=\left(n+1\right).\frac{3}{\left(n+3\right)}=\frac{3\left(n+1\right)}{n+3}\)

Do \(0< n+1< n+3\Rightarrow\frac{n+1}{n+3}< 1\Rightarrow\frac{3\left(n+1\right)}{n+3}< 3\)

Vậy \(A< 3\)

\(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)

\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right)\)

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

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

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

n=1=> đẳng thức đúng

giả sử có số n=a thoả mãn pt=>

2+5+8+....+(3a-1)=a(3a+1)/2=(3a^2+a)/2(1)

phải chứng minh n=a+1 thoả mãn pt:

2+5+8+......+(3a+2)=(a+1)(3a+4)/2=(3a^2+7a+4)/2(2)

lấy (2) trừ (1) ta được:

(6a+4)/2=3a+2

=> 0=0 (đúng vs mọi a)

=> đẳng thức (2) đúg, dpcm

Gọi ĐTV hay lê chí cường ấy

Đặt A = 2 + 5+  ....... + (2n - 1)

Số các số hạng là:

(3n - 1 - 2)/3 + 1 = (3n - 3)/3 + 1 = n - 1 + 1 = n

A = n x (3n -1 + 2) : 2

A = \(\frac{n\left(3n+1\right)}{2}\) => DPCM 

2/3

2/3

2/3 là sao mình ko hiểu lắm