\(A=1.2+2.3+3.4+...+n\left(n+1\right)\)

\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+n\left(n+1\right).3\)

\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n\left(n+1\right).\)\(\left(n+2-n+1\right)\)

\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+n\left(n+1\right)\left(n+2\right)\)\(-\left(n-1\right)n\left(n+1\right)\)

\(\Rightarrow3A=n\left(n+1\right)\left(n+2\right)\)

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

Vì A là số tự nhiên nên A chia hết cho 3 (đpcm)

Ta có: \(\frac{1}{1.2}=\frac{3}{1.2.3}\) ;\(\frac{1}{1.2+2.3}=\frac{3}{2.3.4}\); \(\frac{1}{2.3+3.4}=\frac{3}{3.4.5}\); ......;\(\frac{1}{1.2+2.3+3.4+...+n\left(n+1\right)}=\frac{3}{n\left(n+1\right)\left(n+2\right)}\)

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

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

Ta lại có: \(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}\); \(\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4}\); \(\frac{2}{3.4.5}=\frac{1}{3.4}-\frac{1}{4.5}\);....;\(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

=> \(\frac{2S}{3}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

=> \(\frac{2S}{3}=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)=> \(S=\frac{3}{4}-\frac{3}{2\left(n+1\right)\left(n+2\right)}< \frac{3}{4}\)

=> \(S< \frac{3}{4}\)

Mình nhầm 1 chỗ: \(\frac{1}{1.2+2.3+3.4}=\frac{3}{3.4.5}\)

\(3D_n=1.2.3+2.3.3+3.4.3+...+n\left(n+1\right)3\)

\(=1.2\left(3-0\right)+2.3\left(4-1\right)+...+n\left(n+1\right)\left(n+2-n+1\right)\)

\(=1.2.3-0.1.2+2.3.4-1.2.3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)

\(=n\left(n+1\right)\left(n+2\right)-0.1.2=n\left(n+1\right)\left(n+2\right)\)

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

D_n = 1.2 + 2.3 + 3.4 +...+ n(n + 1)

3D_n = 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) +...+ n(n + 1).[(n + 2) - (n - 1)]

3D_n = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +...+ n(n + 1)(n + 2) - (n - 1)n(n + 1)

3D_n = [1.2.3 + 2.3.4 + 3.4.5 +...+ n(n + 1)(n + 2)] - [0.1.2 + 1.2.3 + 2.3.4 +...+ n(n - 1)(n + 1)]

3D_n = n(n + 1)(n + 2) - 0.1.2

3D_n = n(n + 1)(n + 2)

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

Đặt \(A=1.2+2.3+3.4+...+n\left(n+1\right)\)

\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+3n\left(n+1\right)\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n\left(n+1\right)\left(n+2-n+1\right)\)

\(=1.2.3+2.3.4-1.2.3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)

\(=n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow1.2+2.3+3.4+...+n\left(n+1\right)=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

Bạn ơi tại sao 3n.(n+1) lại bằng với n.(n+1).(n+2-n+1)

A=1.2+2.3+...+n(n+1)

3A=1.2.3+2.3.3+....+3n(n+1)

3A=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+n(n+1)(n+2)-(n-1)n(n+1)

3A=n(n+1)(n+2)

A=n(n+1)(n+2)/3 (đpcm)

Ta có :

\(A=\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+...+\frac{\left(n-1\right)n-1}{n!}\)

\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{\left(n-1\right)n}{n!}-\frac{1}{n!}\)

\(=1-\frac{1}{2!}+1-\frac{1}{3!}+\frac{1}{2!}-\frac{1}{4}!+\frac{1}{3!}-\frac{1}{5!}+\frac{1}{4!}-...+\frac{1}{\left(n-2\right)!}-\frac{1}{n!}\)

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

Vậy ...

A=1.2+2.3+....+n(n+1)

3A=1.2.3+2.3.3+....+3n(n+1)

3A=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+n(n+1)(n+2)-(n-1)n(n+1)

3A=n(n+1)(n+2)

A=n(n+1)(n+2)/3 (đpcm)

- Với \(n=1\Rightarrow1.2=\frac{1.2.3}{3}\) (đúng)

- Giả sử đúng với \(n=k\) hay \(1.2+...+k\left(k+1\right)=\frac{k\left(k+1\right)\left(k+2\right)}{3}\)

Ta cần chứng minh nó đúng với \(n=k+1\) hay:

\(1.2+...+k\left(k+1\right)+\left(k+1\right)\left(k+2\right)=\frac{\left(k+1\right)\left(k+2\right)\left(k+3\right)}{3}\)

Thật vậy:

\(1.2+...+k\left(k+1\right)+\left(k+1\right)\left(k+2\right)\)

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

\(=\left(k+1\right)\left(k+2\right)\left[\frac{k}{3}+1\right]=\frac{\left(k+1\right)\left(k+2\right)\left(k+3\right)}{3}\) (đpcm)