Question

[Lê Viết HIếu]

[\(\frac{2000}{2000.2006}+\frac{2000}{2006.2012}+\frac{2000}{2012.2018}+.....+\frac{2000}{2492.2498}\)]x\(\frac{^{3^2}}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}+.....+\frac{3^2}{197.200}\)

Answer 1

[Đức Phạm]

\(\left[\frac{2000}{2000.2006}+\frac{2000}{2006.2012}+...+\frac{2000}{2492.2498}\right]\times\left[\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}+...+\frac{3^2}{197.200}\right]\)

\(=\left[\frac{2000}{6}\cdot\left(\frac{1}{2000}-\frac{1}{2006}+...+\frac{1}{2492}-\frac{1}{2498}\right)\right]\times\left[\frac{9}{8.11}+\frac{9}{11.14}+...+\frac{9}{197.200}\right]\)

\(=\left[\frac{2000}{6}\cdot\left(\frac{1}{2000}-\frac{1}{2498}\right)\right]\times\left[\frac{9}{3}\cdot\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+..+\frac{1}{197}-\frac{1}{200}\right)\right]\)

\(=\left[\frac{2000}{6}\cdot\frac{498}{4996000}\right]\times\left[\frac{9}{3}\cdot\left(\frac{1}{8}-\frac{1}{200}\right)\right]\)

\(=\frac{83}{2498}\times\left[\frac{9}{3}\cdot\frac{3}{25}\right]\)

\(=\frac{83}{2498}\times\frac{9}{25}=\frac{747}{62450}\)