Sorry đăng làm giwor thì em nó bấm nộp bài mk làm tiếp nhé
\(E=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+......+\frac{1}{1+2+3+.....+24}\)
\(=\frac{1}{\frac{\left(2-1\right).2}{2}}+\frac{1}{\frac{\left(3-1\right).3}{2}}+.....+\frac{1}{\frac{\left(24-1\right).24}{2}}\)
\(=\frac{1}{\frac{1.2}{2}}+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+.....+\frac{1}{\frac{23.24}{2}}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+.....+\frac{2}{23.24}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{23.24}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{23}-\frac{1}{24}\right)\)
\(=2\left(1-\frac{1}{24}\right)\)
\(=2.\frac{23}{24}=\frac{23}{12}\)
Vậy tỉ số giữa D và E là ; \(\frac{5}{28}:\frac{23}{2}=\frac{5}{322}\)
Ta có : \(D=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+.....+\frac{10}{1400}\)
\(=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+.....+\frac{5}{25.28}\)
\(=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+.....+\frac{3}{25.28}\right)\)
\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{25}-\frac{1}{28}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)\)
\(=\frac{5}{3}.\frac{3}{28}=\frac{5}{28}\)
