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

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)

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

\(=1-\frac{1}{50}=\frac{49}{50}\)

\(B=1.2+2.3+3.4+...+49.50\)

\(3B=1.2.3+2.3.3+3.4.3+...+49.50.3\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)

\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)

\(=49.50.51\)

\(B=\frac{49.50.51}{3}=49.50.17\)

\(50^2.A-\frac{B}{17}=49.50-49.50=0\)

Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 32.33

=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 32.33.34

=> 3S = 32.33.34

=> S = \(\frac{32.33.34}{3}=11968\)

Ta thấy:\(\frac{1}{1.2}=1-\frac{1}{2},\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3},...,\frac{1}{49.50}=\frac{1}{49}-\frac{1}{50}\)

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

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

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

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

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

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

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

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

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)

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

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

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

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

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

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

3A=49.50.51

3A=124950

A=41650

A=1.2+2.3+3.4+...+49.50

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

= 1.2.3-1.2.3+2.3.4-2.3.4+...-48.49.50 +49.50.51

=49.50.51

=124950

\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49\cdot50}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ =1-\dfrac{1}{50}\\ =\dfrac{49}{50}\)

A = 1- 1/2 + 1/2-1/3 +1/3-1/4+...........+ 1/49-1/50

A= 1- 1/50= 49/50

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

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

\(A=\frac{1}{1}-0+0+0+0+......+0+0-\frac{1}{50}\)

\(A=\frac{1}{1}-\frac{1}{50}=\frac{49}{50}\)

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

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

3A = 49.50.51

A = 41650

\(A=1.2+2.3+3.4+....+49.50\)

\(3A=1.2.3+2.3.3+3.4.3+....+49.50.3\)

\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+....+49.50.\left(51-48\right)\)

\(3A=1.2.3+2.3.4+3.4.5+....+49.50.51\)

\(3A=49.50.51=124950\)

\(\Rightarrow A=\frac{124950}{3}=41650\)