Question

[Nết Đặng]

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{99\cdot100}\)

Tìm A

Answer 1

[Nguyễn Ngọc Quý]

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

Xét thấy: \(\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2};\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3};.....\)

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

Answer 2

[Hoàng Phúc]

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

=1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100

=1/1-1/100=100/100-1/100=99/100

 Vậy A=99/100