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

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

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

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

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

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

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

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

\(=49.50.51\)

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

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

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 = \(\frac{32.33.34}{3}=11968\)

41650 tk m nhé

Nhân cả 2 vế của S với 3 ta được :

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

= 1.2.3 + 2.3.3 + 3.4.3 + ... + 49.50.3

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

= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 49.50.51 - 48.59.50

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

= 49.50.51

=> S = 49.50.51/3 = 41650

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

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

=49.50.51

=124950

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Đặt tổng trên =A

\(3A=1.2.3+2.3.3+3.4.3+...+48.49+49.50\)

\(3A=1.2.3+2.3\left(4-1\right)+3.4.\left(5-2\right)+...+48.49\left(50-47\right)+49.50\left(51-48\right)\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+48.49.50-47.48.49+49.50.51-48.49.50\)

\(3A=49.50.51\Rightarrow A=\frac{49.50.51}{3}=17.50.51\)

đặt A=1.2+2.3+...+49.50

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

3A=1.2.(3-0)+2.3.(4-1)+...+49.50.(51-48)

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

3A=49.50.51

A=\(\frac{49.50.51}{3}\)

=>A=41650

 Dãy số trên lập thành dãy số có khoảng cách là 1 đơn vị .

Số lượng số của dãy trên là :

          ( 49.50 - 1.2 ) : 1 x 1 = 49.3

Tổng của dãy số trên là :

        ( 49.50 + 1.2 ) x 49.3 :2 = 1249.755

                                           Đ/S: 1249.755

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

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

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

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

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

\(3A=49.50.51=124950\)

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

\(\frac{2.4+4.6+6.8+...+98.100}{1.2+2.3+3.4+...+49.50}=\frac{4.\left(1.2+2.3+3.4+...+49.50\right)}{1.2+2.3+3.4+...+49.50}=\frac{4}{1}=4\)

(0,5+0,5)÷0,5_0,5×0,5=

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