Question

B=\(\dfrac{4}{1.4}\)+\(\dfrac{4}{4.7}\)+\(\dfrac{4}{7.10}\)+....+\(\dfrac{4}{100.103}\)+\(\dfrac{4}{103.106}\)

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Answer 1

\(B=\dfrac{4}{1.4}+\dfrac{4}{4.7}+\dfrac{4}{7.10}+...+\dfrac{4}{100.103}+\dfrac{4}{103.106}\)

\(B=4\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}+\dfrac{1}{103.106}\right)\)

\(B=\dfrac{4}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}+\dfrac{3}{103.106}\right)\)

\(B=\dfrac{4}{3}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}+\dfrac{1}{103}-\dfrac{1}{106}\right)\)

\(B=\dfrac{4}{3}\left(\dfrac{1}{3}-\dfrac{1}{106}\right)\)

\(B=\dfrac{4}{3}.\dfrac{103}{318}\)

\(B=\dfrac{412}{954}\)