Question

Tính : 

A, 1/1.2 + 1/ 2.3 + 1/3.4 + ....+1/99.100

B, 4/1.5 + 4/ 5.9+4/9.13 + ...... + 4/ (n-4).n    ( n khác 4 , n khác 0 )

Mk sẽ tick cho

Answer 1

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

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

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

Answer 2

Còn câu b niwax nha các bn . Giúp mk với