e: \(C=\dfrac{1}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{97\cdot100}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{99}{100}=\dfrac{33}{100}\)
e) \(C=\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}\)
\(\Rightarrow3C=\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\)
\(\Rightarrow3C=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
\(\Rightarrow3C=1-\dfrac{1}{100}=\dfrac{99}{100}\Rightarrow C=\dfrac{33}{100}\)
f) \(\dfrac{4}{1.3.5}+\dfrac{4}{3.5.7}+...+\dfrac{4}{95.97.99}\)
\(=\dfrac{5-1}{1.3.5}+\dfrac{7-3}{3.5.7}+...+\dfrac{99-95}{95.97.99}\)
\(=\dfrac{4}{1.3}-\dfrac{4}{3.5}+\dfrac{4}{3.5}-\dfrac{4}{5.7}+...+\dfrac{4}{95.97}-\dfrac{4}{97.99}\)
\(=1-\dfrac{4}{97.99}=\dfrac{9599}{9603}\)
