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A = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}}{\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{48}{2}+\frac{49}{1}}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}}{\left(\frac{1}{49}+1\right)+\left(\frac{2}{48}+1\right)+\left(\frac{3}{47}+1\right)+...+\left(\frac{48}{2}+1\right)+\frac{50}{50}}\)
A = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}}{\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+...+\frac{50}{2}+\frac{50}{50}}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}}{\left(\frac{1}{49}+\frac{1}{48}+\frac{50}{47}+...+\frac{1}{2}+\frac{1}{50}\right).50}=\frac{1}{50}\)
\(A=\frac{T}{M}\)
\(M=\frac{1}{49}+1+\frac{2}{48}+1+\frac{3}{47}+1+.........+\frac{48}{2}+1+1\)
\(=\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+.........+\frac{50}{2}+1\)
\(=50.\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+......+\frac{1}{2}+\frac{1}{50}\right)=50.T\)
\(A=\frac{T}{50T}=\frac{1}{50}\)
Đặt \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{2}{2004.2005}\)
Ta có: \(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{2004.2005}\)
\(A=2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{2004.2005}\right)\)
\(A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{2004.2005}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{2005}\right)\)
\(A=\frac{2003}{2005}\)
\(A=3+3^2+...+3^{50}\)
\(\Rightarrow3A=3^2+3^3+...+3^{50}+3^{51}\)
\(\Rightarrow3A-A=3^{51}-3\)
\(\Rightarrow2A=3^{51}-3\)
\(\Rightarrow A=\frac{3^{51}-3}{2}\)
\(B=2-2^2+2^3-2^4+...+2^{2019}-2^{2020}\)
\(2B=2^2-2^3+2^4-2^5+...+2^{2020}-2^{2021}\)
\(B+2B=2-2^{2021}\)
\(3B=2-2^{2021}\)
\(B=\frac{2-2^{2021}}{3}\)
\(C=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2008.2009}\)
\(C=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2008}-\frac{1}{2009}\)
\(C=1-\frac{1}{2009}\)
\(C=\frac{2008}{2009}\)
\(D=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(D=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(D=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(D=\frac{1}{2}\left(1-\frac{1}{11}\right)\)
\(D=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)