\(E=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{200}\left(1+2+...+199+200\right)\)
\(=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{200}\cdot\dfrac{200\cdot201}{2}\)
\(=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{201}{2}\)
\(=\dfrac{2+3+...+201}{2}=\dfrac{\dfrac{\left(201+2\right)\left(201-2+1\right)}{2}}{2}=203\cdot\dfrac{200}{4}=203\cdot50\)
\(\dfrac{E}{F}=\dfrac{203\cdot50}{\dfrac{20300}{3}}=203\cdot50\cdot\dfrac{3}{20300}=50\cdot\dfrac{3}{100}=\dfrac{3}{2}\)
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