Trả lời :

A = 1 . 2 + 2 . 3 + 3 . 4 + ... + 30 . 31

=> 3A = 1 . 2 . 3 + 2 . 3 . 3 + 3 . 4 . 3 + ... + 30 . 31 . 3

=> 3A = 1 . 2 . 3 + 2 . 3 . (4 - 1) + 3 . 4 . (5 - 2) + ... + 30 . 31 . (32 - 29)

=> 3A = 1 . 2 . 3 + 2 . 3 . 4 - 1 . 2 . 3 + 3 . 4 . 5 - 2 . 3 . 4 + ... + 30 . 31 . 32 - 29 . 30 . 31

=> 3A = 30 . 31 . 32

=> 3A = 29760

=> A = 9920

a) \(A=1.2+2.3+3.4+.........+30.31\)

\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+........+30.31.3\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+.....+30.31.\left(32-29\right)\)

\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+......+30.31.32-29.30.31\)

\(=30.31.32\)

\(\Rightarrow A=\frac{30.31.32}{3}=9920\)

b) \(B=1+\left(1+2\right)+\left(1+2+3\right)+........+\left(1+2+...........+20\right)\)

\(=\frac{1.2}{2}+\frac{2.3}{2}+\frac{3.4}{2}+.......+\frac{20.21}{2}\)

\(=\frac{1.2+2.3+3.4+.......+20.21}{2}\)

Làm tương tự như phần a ta được: 

\(1.2+2.3+3.4+.......+20.21=\frac{20.21.22}{3}=3080\)

\(\Rightarrow B=\frac{3080}{2}=1540\)

các bạn giúp mk với

A = 1.2 + 2.3 + 3.4 + ... + 99.100 + 100.101

⇒ 3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3 + 100.101.3

= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98) + 100.101.(102 - 99)

= 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + 3.4.5 + ... - 98.99.100 + 99.100.101 - 99.100.101 + 100.101.102

= 100.101.102

= 1030200

⇒ A = 1030200 : 3

= 343400

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Ta có :

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{1}-\frac{1}{100}\)

\(A=\frac{100}{100}-\frac{1}{100}\)

\(A=\frac{99}{100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

Ai k mk mk k lại

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

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

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

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

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

\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}+\frac{51-50}{50.51}\)

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

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

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

A= 1/1.2 +1/2.3+1/3.4+ .........1/49.50

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

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

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

=>\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{10}{11}\)

=>\(1-\frac{1}{x+1}=\frac{10}{11}\)

=>\(\frac{1}{x+1}=1-\frac{10}{11}\)

=>\(\frac{1}{x+1}=\frac{1}{11}\)

=>x+1=11

=>x=10