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\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}\)
\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\left[\frac{1}{199}+1\right]+\left[\frac{2}{198}+1\right]+\left[\frac{3}{197}+1\right]+...+\left[\frac{198}{2}+1\right]}\)
\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}}\)
\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{200\left[\frac{1}{199}+\frac{1}{198}+\frac{1}{197}+...+\frac{1}{2}\right]}=\frac{1}{200}\)
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B = \(\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}\)
\(=\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+199\)
\(=\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+...+\left(\frac{198}{2}+1\right)+1\)
(từ 1 đến 198 có 198 số hạng nên còn 1 số 1)
\(=\frac{200}{199}+\frac{200}{198}+...\frac{200}{2}+\frac{200}{200}\)
\(=200\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{\text{4}}+...+\frac{1}{200}\right)=200A\)
=> B = 200A => \(\frac{A}{B}=\frac{1}{200}\)
Vậy \(\frac{A}{B}=\frac{1}{200}\)
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\(B=\frac{1}{199}+\frac{2}{198}+...+\frac{199}{1}\)
\(=1+\frac{1}{199}+1+\frac{2}{198}+...+\frac{199}{1}+1-199\)
\(=200+\frac{200}{2}+...+\frac{200}{199}-199\)
\(=1+\frac{200}{2}+...+\frac{200}{199}\)
\(=200\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{200\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)}=\frac{1}{200}\)
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Sửa đề \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}\)
\(B=\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}\)
\(=\left(1+\frac{1}{199}\right)+\left(\frac{2}{198}+1\right)+\left(\frac{3}{197}+1\right)+...+\left(\frac{2}{198}+1\right)+1\)
\(=\frac{200}{200}+\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}\)
\(=200\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)
Khi đó A/B = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{200\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)}=\frac{1}{200}\)
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a, Dat A =\(\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-...-\frac{1}{3^{198}}+\frac{1}{3^{199}}\)
\(\Rightarrow\frac{1}{3}A=\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...-\frac{1}{3^{199}}+\frac{1}{3^{200}}\)
\(\Rightarrow\frac{1}{3}A+A=\left(\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...-\frac{1}{3^{199}}+\frac{1}{3^{200}}\right)+\left(\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-...-\frac{1}{3^{198}}+\frac{1}{3^{199}}\right)\)
\(\Rightarrow\frac{4}{3}A=\frac{1}{3}+\frac{1}{3^{200}}\)
\(\Rightarrow A=\frac{\frac{1}{3}+\frac{1}{3^{200}}}{\frac{4}{3}}\)
chung minh tuong tu cau b va c