\(B=\dfrac{5}{1.4}+\dfrac{5}{4.7}+.....+\dfrac{5}{100.103}\)
\(B=\dfrac{5}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\right)\)
\(B=\dfrac{5}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(B=\dfrac{5}{3}.\left(1-\dfrac{1}{103}\right)\)
\(B=\dfrac{5}{3}.\dfrac{102}{103}\)
\(B=1\dfrac{67}{103}\)
Ta có: \(B=\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)
\(=\dfrac{5}{3}\left(1-\dfrac{1}{4}\right)+\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}\right)+...+\dfrac{5}{3}\left(\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(=\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(=\dfrac{5}{3}\left(1-\dfrac{1}{103}\right)\)
\(=\dfrac{5}{3}.\dfrac{102}{103}=\dfrac{170}{103}\)
Vậy \(B=\dfrac{170}{103}\).
Ta có : B = \(\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)
\(\Rightarrow B=\dfrac{5}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\right)\)
\(\Rightarrow B=\dfrac{5}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(\Rightarrow B=\dfrac{5}{3}.\left(1-\dfrac{1}{103}\right)\)
\(\Rightarrow B=\dfrac{5}{3}.\dfrac{102}{103}\)
\(\Rightarrow B=\dfrac{170}{103}\)
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