Question

Tính \(\dfrac{A}{B}\) biết rằng:

A = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{200}\)

B = \(\dfrac{1}{199}+\dfrac{2}{198}+\dfrac{3}{197}+...+\dfrac{198}{2}+\dfrac{199}{1}\)

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Answer 1

Ta có :

\(\dfrac{1}{199}+\dfrac{2}{198}+...+\dfrac{198}{2}+\dfrac{199}{1}\)

\(=\left(\dfrac{1}{199}+1\right)+\left(\dfrac{2}{198}+1\right)+...+\left(\dfrac{198}{2}+1\right)\left(\dfrac{199}{1}+1\right)-199\)\(=\dfrac{200}{199}+\dfrac{200}{199}+...+\dfrac{200}{2}+200-199\)

\(=\dfrac{200}{199}+\dfrac{200}{198}+...+\dfrac{200}{2}+\dfrac{200}{200}\)

\(=200\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{200}\right)\)

\(=200.A\)

\(\Rightarrow\dfrac{A}{B}=\dfrac{1}{200}\)