\(B=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{2014}-\frac{1}{2017}\right)\)
\(B=\frac{5}{3}.\left(1-\frac{1}{2017}\right)\)
\(B=\frac{5}{3}.\frac{2016}{2017}=\frac{10080}{6051}\)
\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{2014.2017}\)
\(3M=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{2014.2017}\right)\)
\(3M=5\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2014}-\frac{1}{2017}\right)\)
\(3M=5\left(1-\frac{1}{2017}\right)\)
\(3M=5.\frac{2016}{2017}\)
\(3M=\frac{10080}{2017}\)
\(\Rightarrow M=\frac{3360}{2017}\)
Ta có:
\(B=\frac{5}{1.4}+\frac{5}{4.7}+.....+\frac{5}{2014.2017}\)
\(\Rightarrow B.\frac{3}{5}=\frac{3}{1.4}+\frac{3}{4.7}+.....+\frac{3}{2014.2017}\)
\(=\left(1-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+......+\left(\frac{1}{2014}-\frac{1}{2017}\right)\)
\(=1-\frac{1}{2017}\)
\(=\frac{2016}{2017}\)
\(\Rightarrow B=\frac{2016}{2017}:\frac{3}{5}\)
\(=\frac{3360}{2017}\)
Vậy B \(=\frac{3360}{2017}\)~~~~~
\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{2014.2017}\)
\(\Rightarrow B=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{2014.2017}\right)\)
\(\Rightarrow3B=5\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{2014.2017}\right)\)
\(\Rightarrow3B=5\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2014}-\frac{1}{2017}\right)\)
\(\Rightarrow3B=5\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow3B=5.\frac{2016}{2017}\)
\(\Rightarrow B=\frac{5.2016\div3}{2017}\)
\(\Rightarrow B=\frac{3360}{2017}\)