Ta có
\(A=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(=\frac{7}{4}.\left[\frac{3333}{101}.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\right]\)
\(=\frac{7}{4}.\left[\frac{3333}{101}.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\right]\)
\(=\frac{7}{4}.\left[\frac{3333}{101}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\right]\)
\(=\frac{7}{4}.\left[\frac{3333}{101}.\left(\frac{1}{3}-\frac{1}{7}\right)\right]\)
\(=\frac{7}{4}.\frac{3333}{101}.\frac{4}{21}=\frac{1111}{101}\)
bạn nguyễn văn kiệt làm nhầm rồi
