Đề sai rồi Sửa lại thành \(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{1023}\) nha
\(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{1023}\)
B = \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{31\cdot33}\)
B = \(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{33}\)
B = \(1-\dfrac{1}{33}\)
B = \(\dfrac{32}{33}\)