Question

A=\(\frac{1}{1.3}\)+\(\frac{1}{3.5}\)+.....+\(\frac{1}{1999.2001}\)

B=\(\frac{1}{3.5}\)+\(\frac{1}{5.7}\)+.....+\(\frac{1}{99.101}\)

Answer 1

A = 1/1.3 + 1/3.5 + 1/5.7 +........+ 1/1999.2001
2.A = 2/1.3 + 2/3.5 + 2/5.7 +........+ 2/1999.2001
2.A = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/1999 - 1/2001 
2.A = 1 - 1/2001 

2.A = 2000/2001

Vậy A =1000/2001

B = 1/3.5 + 1/5.7 + 1/7.9 +........+ 1/99.101
2.A = 2/3.5 + 2/5.7 + 2/7.9 +........+ 2/99.101
2.A = 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/99 - 1/101 
2.A = 1/3 - 1/101 = 98/303 
Vậy A =49/303

Answer 2

\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{1999.2001}\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{1999.2001}\)

\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{1999}-\frac{1}{2001}\)

\(2A=\frac{1}{1}-\frac{1}{2001}=\frac{2000}{2001}\)

\(A=\frac{2000}{2001}.\frac{1}{2}=\frac{1000}{2001}\)