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Đặt \(A=\frac{7}{1\cdot2}+\frac{7}{2\cdot3}+...+\frac{7}{10\cdot11}\)
\(\Rightarrow\frac{1}{7}A=\frac{1}{7}\left(\frac{7}{1\cdot2}+\frac{7}{2\cdot3}+...+\frac{7}{10\cdot11}\right)\)
\(\Rightarrow\frac{1}{7}A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{10\cdot11}\)
\(\Rightarrow\frac{1}{7}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)
\(\Rightarrow\frac{1}{7}A=1-\frac{1}{11}\)
\(\Rightarrow\frac{1}{7}A=\frac{10}{11}\)
\(\Rightarrow A=\frac{10}{11}:\frac{1}{7}\)
\(\Rightarrow A=\frac{70}{11}\)
Chị dùg cách tính tổng đi
1. Tìm dãy cách đều bao nhiêu
2. Từ công thức tính tổng rồi suy ra
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{100.101}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{100}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)\(=\frac{101}{101}-\frac{1}{101}\)
\(=\frac{100}{101}\)
Ta có \(\frac{-17}{30}-\frac{-11}{15}-\frac{7}{12}\)
\(=\frac{-17}{30}+\frac{11}{15}-\frac{7}{12}\)
\(=\left(\frac{-17}{30}+\frac{22}{30}\right)-\frac{7}{12}\)
\(=\frac{1}{6}-\frac{7}{12}=\frac{2}{12}-\frac{7}{12}=-\frac{5}{12}\)
Đặt A= 1.2+2.3 +.......+99.100
3A= 1.2.3+2.3.4+3.4.3 +......+ 99.100.3
3A= 1.2. (3 - 0) + 2.3.(4 - 1) +3.4. (5 - 2)....... . 99.100. (101 - 98)
3A = (1.2.3 + 2.3.4 + 3.4.5 +...... + 99.100.101) - (0.1.2 + 1.2.3 + 2.3.4 +.......+ 98.99.100)
3A = 99.100.101 - 0.1.2
3A = 999900 - 0
3A= 999900
A= 999900 : 3
A = 333300
A=1.2+2.3+3.4+…+99.100
3A = 1.2.3 + 2.3.3 + ... + 99.100.3
3A = 1.2.3 + 2.3.(4-1) + ... + 99.100.(101-98)
3A = 1.2.3 + 2.3.4 - 1.2.3 + ... + 99.100.101 - 98.99.100
3A = 99.100.101
=> A = \(\frac{99.100.101}{3}\)= 333 300
Đặt A= 1.2+2.3 +.......+99.100
3A= 1.2.3+2.3.4+3.4.3 +......+ 99.100.3
3A= 1.2. (3 - 0) + 2.3.(4 - 1) +3.4. (5 - 2)....... . 99.100. (101 - 98)
3A = (1.2.3 + 2.3.4 + 3.4.5 +...... + 99.100.101) - (0.1.2 + 1.2.3 + 2.3.4 +.......+ 98.99.100)
3A = 99.100.101 - 0.1.2
3A = 999900 - 0
3A= 999900
A= 999900 : 3
A = 333300
Đặt A = 1.2+2.3 +.......+99.100
=> 3A = 3.( 1.2+2.3 +.......+99.100 )
=> 3A = 1.2. (3 - 0) + 2.3.(4 - 1) +3.4. (5 - 2)....... . 99.100. (101 - 98)
=> 3A = 0.1.2 + 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ..... + 99.100.101
=> 3A =0.1.2 + 99.100.101
=> 3A = 999900
=> A = 999900 : 3
Vậy A = 333300
=\(11\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99\cdot100}\right)\)=\(11\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)=\(11\left(1-\frac{1}{100}\right)\)=11\(\frac{99}{100}\)=\(\frac{1089}{100}\)
Đặt \(A=\frac{11}{1.2}+\frac{11}{2.3}+...+\frac{11}{99.100}\)
\(\Rightarrow A=11\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow A=11\left(1-\frac{1}{100}\right)\)
\(\Rightarrow A=11.\frac{99}{100}\)
\(\Rightarrow A=\frac{1089}{100}\)