Question

[Lan Phạm]

a) 1/1.2+1/2.3+1/3.4+...+1/99.100

b) 2/3.5+2/5.7+...+2/49.51

Answer 1

[Trần Thanh Phương]

a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)

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

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

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

b) \(\frac{2}{3\cdot5}+\frac{3}{5\cdot7}+...+\frac{2}{49\cdot51}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)

\(=\frac{1}{3}-\frac{1}{51}\)

\(=\frac{16}{51}\)

Answer 2

[Đào Trần Tuấn Anh]

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

= 99/100

b) 2/3.5+2/5.7+...+2/49.51

= 2 . ( 1/3.5 + 1/5.7 + ... + 1/49.51 )

= 2 . ( 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/49 - 1/50 )

= 2 . ( 1/3 - 1/50 )

= 2 . 47/150

= 47/75