Question

[Phan Minh Thiện]

\(1\frac{1}{6}+1\frac{1}{12}+1\frac{1}{20}+1\frac{1}{30}+...1\frac{1}{9900}=?\)

Answer 1

[Cô nàng cự giải]

Đặt A = \(1\frac{1}{6}+1\frac{1}{12}+1\frac{1}{20}+1\frac{1}{30}+...+1\frac{1}{9900}\)

=> A = \(1+\frac{1}{6}+1+\frac{1}{12}+1+\frac{1}{20}+1+\frac{1}{30}+...+1+\frac{1}{9900}\)

=> A = \(\left(1+1+1+1+...+1\right)+\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\right)\)

=> A = \(\left(1+1+1+1+...+1\right)+\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\right)\)

=> A = \(99+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\right)\)

=> A = \(99+\left(\frac{1}{2}-\frac{1}{100}\right)\)

=> A = \(99+\frac{49}{100}\)

=> A = \(\frac{9949}{100}\)