Chào bạn, bạn hãy theo dõi bài giải của mình nhé!
Ta có :
\(A=\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+...+\frac{3^2}{340}\)
\(=>A=\frac{9}{2\cdot5}+\frac{9}{5\cdot8}+\frac{9}{8\cdot11}+...+\frac{9}{17\cdot20}\)
\(=>A=\frac{9}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{17\cdot20}\right)\)
\(=>A=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(=>A=3\left(\frac{1}{2}-\frac{1}{20}\right)=3\left(\frac{10}{20}-\frac{1}{20}\right)=3\cdot\frac{9}{20}=\frac{27}{20}\)
Chúc bạn học tốt!
Chọn mình nhé ! 
Ta có:
\(A=\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+...+\frac{3^2}{340}\)
\(\Rightarrow A=3\left(\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{340}\right)\)
\(\Leftrightarrow A=3\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\right)\)
\(\Leftrightarrow A=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(\Leftrightarrow A=3\left(\frac{1}{2}-\frac{1}{20}\right)=3.\frac{9}{20}=\frac{27}{20}\)
Vậy \(A=\frac{27}{20}\)
