Ta có: 3S = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2)  + .....+ 50.51.(52 -49) 

              = 1.2.3 - 0  + 2.3.4 - 1.2.3 + 3.4.5 -2.3.4 + .....+ 50.51.52 - 49.50.51

 3S = 50.51.52

  S = 50.17.52 =44200

Ta có:A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+......+\dfrac{1}{49.50}+\dfrac{1}{50.51}\)

A=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.......+\dfrac{1}{49}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{51}\)

A=1-\(\dfrac{1}{51}=\dfrac{50}{51}\)

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}+\dfrac{1}{50.51}\)

\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{51}\)

\(A=\dfrac{1}{1}-\dfrac{1}{51}\)

\(A=\dfrac{50}{51}\)

A = \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}+\dfrac{1}{50.51}\)

= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{51}\)

= \(1-\dfrac{1}{51}\)

= \(\dfrac{50}{51}\)

 A=1.2+2.3+3.4+...+49.50+50.51

3A= 1.2.3+2.3.3+3.4.3+...+49.50.3+50.51.3

3A= 1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+49.50.(51-48)+50.51.(52-49)

3A= 0.1.2 - 1.2.3 + 1.2.3- 2.3.4 + 2.3.4 - 3.4.5 + ... + 48.49.50 - 49.50.51 + 49.50.51 - 50.51.52

3A= 50.51.52

3A=132600

  A=66300

\(A=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}=\frac{1}{1}-\frac{1}{50}=\frac{49}{50}\)

Vậy A=49/50

Công thức: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

A=1.2+2.3+...+49.50

3A=1.2.3+2.3.3+...+49.50.3

3A=1.2.(4-1)+2.3.(5-2)+....+49.50.(51-48)

3A=1.2.4-1.2.1+2.3.5-2.3.2+...+49.50.51-49.50.48

3A=49.50.51

=>A=49.25.51

=>A=62475

A=1.2+2.3+...+49.50

3A=1.2.3+2.3.3+...+49.50.3

3A=1.2.(4-1)+2.3.(5-2)+....+49.50.(51-48)

3A=1.2.4-1.2.1+2.3.5-2.3.2+...+49.50.51-49.50.48

3A=49.50.51

=>A=49.25.51

=>A=62475

3S=(1.2+2.3+3.4+...+49.50).3

3S=1.2.3+2.3.3+3.4.3+...+49.50.3

3S=1.2.3+2.3.(4-1)+3.4.(5-2)+...+49.50.(51-48)

3S=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50

3S=49.50.51

S=17.49.50

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}< 1\) (đpcm)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}\)

\(=\frac{49}{50}\)

\(\Rightarrow\) Quy đồng phân số và 1 là : \(\frac{49}{50}\) và \(1\)

Giữ nguyên phân số \(\frac{49}{50}\)

Ta có : \(\frac{1}{1}=\frac{1.50}{1.50}=\frac{50}{50}\)

\(\Rightarrow\frac{49}{50}< \frac{50}{50}\left(đpcm\right)\)