Question

\(E=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{98\cdot99\cdot100}\)

Answer 1

\(E=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\)

\(E=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\)

\(E=\dfrac{1}{1.2}-\dfrac{1}{99.100}=\dfrac{4949}{9900}\)

Answer 2

\(E=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{98.99.100}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{99.100}\right)\)

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Answer 3

E = \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\)

2E = \(\dfrac{2}{1.2.3}+\dfrac{2}{3.4.5}+...+\dfrac{2}{98.99.100}\)

= \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}\right)\)+\(\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)\) +...+ \(\left(\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

= \(\dfrac{1}{1.2}\) - \(\dfrac{1}{99.100}\)= \(\dfrac{1}{2}-\dfrac{1}{9900}\) = \(\dfrac{4949}{9900}\)