Hoàng Việt Nguyễn
10/03/2017 at 18:01-
Faded 26/01/2018 at 12:311.3+13.5+15.7+......+11999.2001
=12(1−13+13−15+.......+11999−12001)
=12(1−12001)
=12.20002001
=10002001
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Phạm Hữu Đang 10/03/2017 at 18:08\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{1999.2001}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{1999}-\dfrac{1}{2001}\)
\(=1-\dfrac{1}{2001}\)
\(=\dfrac{2000}{2001}\)
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Vũ Thị Hương Giang 10/03/2017 at 18:28\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+......+\dfrac{1}{1999.2001}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+.......+\dfrac{1}{1999}-\dfrac{1}{2001}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{2001}\right)\)
\(=\dfrac{1}{2}.\dfrac{2000}{2001}\)
\(=\dfrac{1000}{2001}\)
