\(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{100\cdot103}=\frac{1}{3}\left[\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right]\)
\(\frac{1}{3}\left[1-\frac{1}{103}\right]=\frac{1}{3}\cdot\frac{102}{103}=\frac{34}{103}\)
\(\frac{1}{1\cdot4}\)+\(\frac{1}{4\cdot7}\)+\(\frac{1}{7\cdot10}\)+...+\(\frac{1}{100\cdot103}\)
\(B=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{100.103}\)
\(B=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+..+\frac{103-100}{100.103}\)
\(B3=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{100}-\frac{1}{103}\)
\(B3=\frac{1}{1}-\frac{1}{103}\)
\(B3=\frac{102}{103}\)
\(\Rightarrow B=\frac{102}{103}\times\frac{1}{3}=\frac{34}{103}\)
