Question

Tính tổng 

C=(1-1/2)+(1-1/6)+(1-1/12)+......+(1-1/9900)

 

Answer 1

 C = \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{9900}\right)\)(99 CẶP)

     = \(\left(1+1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\right)\)(99 SỐ HẠNG 1)

     = \(1.99-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

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

     = \(99-\left(1-\frac{1}{100}\right)\)

     = \(99-\frac{99}{100}\)

     = \(\frac{9801}{100}\)

Vậy \(C=\frac{9801}{100}\)

Chúc bạn học tốt !!!!!