Trả lời
\(B=\frac{5}{1\cdot4}+\frac{5}{4\cdot7}+...+\frac{5}{100\cdot103}\)
\(\frac{3}{5}B=\frac{3}{5}\left(\frac{5}{1\cdot4}+\frac{5}{4\cdot7}+...+\frac{5}{100\cdot103}\right)\)
\(\frac{3}{5}B=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{...3}{100\cdot103}\)
\(\frac{3}{5}B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)
\(\frac{3}{5}B=1-\frac{1}{103}\)
\(\frac{3}{5}B=\frac{102}{103}\)
\(B=\frac{102}{103}:\frac{3}{5}\)
\(B=\frac{170}{103}\)
\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
\(B=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{100.103}\right)\)
\(3B=15\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(3B=15\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(3B=15\left(\frac{1}{1}-\frac{1}{100}\right)=15\left(\frac{100}{100}-\frac{1}{100}\right)=15.\frac{99}{100}\)
\(B=\frac{1}{3}.15-\frac{1}{3}.\frac{99}{100}=5-\frac{33}{100}=\frac{500}{100}-\frac{33}{100}=\frac{467}{100}\)
\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
\(=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(=\frac{5}{3}.\left(1-\frac{1}{103}\right)\)
\(=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)
B=\(\frac{5}{1.4}=\frac{5}{4.7}+....+\frac{5}{100.103}\)
= \(\frac{5}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
= \(\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
= \(\frac{5}{3}\left(1-\frac{1}{103}\right)\)
= \(\frac{5}{3}\times\frac{102}{103}\)
= \(\frac{170}{103}\)
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