Dấu \(.\)là dấu nhân
Ta có :
\(E=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{100.103}\)
\(\Rightarrow E=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{2}{100.103}\right)\)
\(\Rightarrow E=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(\Rightarrow E=\frac{2}{3}.\left(1-\frac{1}{103}\right)\)
\(\Rightarrow E=\frac{2}{3}.\frac{102}{103}\)
\(\Rightarrow E=\frac{68}{103}\)
Vậy \(E=\frac{68}{103}\)
~ Ủng hộ nhé
\(E=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+...+\frac{2}{100\cdot103}\)
\(E=2\cdot\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{100\cdot103}\right)\)
Gọi tổng trong ngoặc là F
\(\Rightarrow3F=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{100\cdot103}\)
\(\Rightarrow3F=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)
\(\Rightarrow3F=1-\frac{1}{103}=\frac{102}{103}\)
\(\Rightarrow F=\frac{102}{103\cdot3}=\frac{34}{103}\)
\(\Leftrightarrow E=2\cdot\frac{34}{103}=\frac{68}{103}\)
Vậy......
E = 2/1×4 + 2/4×7 + 2/7×10 + ... + 2/100×103
E = 2( 1/1×4 + 1/4×7 + 1/7×10 + ... + 1/100×103)
3E = 2( 3/1×4 + 3/4×7 + 3/7×10 + ... + 3/100×103 )
3E = 2(1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ...+ 1/100-1/103)
3E = 2( 1 - 1/103 )
3E = 2 × 102/103
3E = 204/103
=> E = 204/103 : 3
E = 68/103
Hok tốt
Ta có \(E=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+.....+\frac{2}{100.103}\)
\(\Leftrightarrow E=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+.....+\frac{3}{100.103}\right)\)
\(\Leftrightarrow E=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{100}-\frac{1}{103}\right)\)
\(\Leftrightarrow E=\frac{2}{3}\left(1-\frac{1}{103}\right)\)
\(\Leftrightarrow E=\frac{2}{3}.\frac{102}{103}=\frac{68}{103}\)