Question

[Vương Hải Nam]

\(\frac{3^2}{8.11}\) +\(\frac{3^2}{11.14}\)+\(\frac{3^2}{14.17}\)+...+\(\frac{3^2}{197.200}\)

Answer 1

[Kiệt Nguyễn]

Đặt \(A=\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}+...+\frac{3^2}{197.200}\)

\(\Leftrightarrow A=\frac{9}{8.11}+\frac{9}{11.14}+\frac{9}{14.17}+...+\frac{9}{197.200}\)

\(\Leftrightarrow\frac{1}{3}A=\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+...+\frac{3}{197.200}\)

​\(\Leftrightarrow\frac{1}{3}A=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{2}{17}+...+\frac{1}{197}-\frac{1}{200}\)​b

\(\Leftrightarrow\frac{1}{3}A=\frac{1}{8}-\frac{1}{200}\)

\(\Leftrightarrow\frac{1}{3}A=\frac{24}{200}\)

\(\Leftrightarrow A=\frac{24}{200}\times3\)

\(\Leftrightarrow A=\frac{72}{200}=\frac{9}{25}\)

Answer 2

[Vương Hải Nam]

Thank