Question

1/7+1/91+...+1/1147

Answer 1

\(\dfrac{1}{7}+\dfrac{1}{91}+...+\dfrac{1}{1147}\)

\(=\dfrac{1}{1\cdot7}+\dfrac{1}{7\cdot13}+\dfrac{1}{13\cdot19}+...+\dfrac{1}{31\cdot37}\)

\(=\dfrac{1}{6}\cdot\left(\dfrac{6}{1\cdot7}+\dfrac{6}{7\cdot13}+\dfrac{6}{13\cdot19}+...+\dfrac{6}{31\cdot37}\right)\)

\(=\dfrac{1}{6}\cdot\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+...+\dfrac{1}{31}-\dfrac{1}{37}\right)\)

\(=\dfrac{1}{6}\cdot\left(1-\dfrac{1}{37}\right)\)

\(=\dfrac{1}{6}\cdot\dfrac{36}{37}\)

\(=\dfrac{6}{37}\)

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#\(Toru\)