Đặt A = \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{31.33}\)
=> 2A = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{31.33}\)
=> 2A = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{31}-\frac{1}{33}\)
=> 2A = \(\frac{1}{3}-\frac{1}{33}\)
=> A = \(\left(\frac{1}{3}-\frac{1}{33}\right):2\)