\(\frac{4}{3\cdot5}+\frac{6}{5\cdot7}+\frac{8}{7\cdot9}+....+\frac{100}{99\cdot101}\)
\(=2\left(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+....+\frac{1}{99\cdot100}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{99}-\frac{1}{101}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(=2\cdot\frac{98}{101}\)
\(=\frac{196}{101}\)
Đúng 0
Bình luận (0)
