Hoàng Việt Nguyễn
10/03/2017 at 18:03-
Faded 26/01/2018 at 12:3011.4+14.7+17.10+...+1100.103
=1−14+14−17+17−110+...+1100−1103
=1−1103
=102103
We have : 11.4+14.7+17.10+.......+1100.103
13(31.4+34.7+37.10+.......+3100.103)
=13(1−14+14−17+.......+1100−1103)
=13(1−1103)
=13.102103=34103
11.4+14.7+17.10+...+1100.103=13(31.4+34.7+37.10+...+1100.103)
=13(11−14+14−17+17−110+...+1100−1103)
=13(1−1103)
=13.102103
=34103
-
Phạm Hữu Đang 10/03/2017 at 18:06\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(=1-\dfrac{1}{103}\)
\(=\dfrac{102}{103}\)
-
Vũ Thị Hương Giang 10/03/2017 at 18:21We have : \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+.......+\dfrac{1}{100.103}\)
\(\dfrac{1}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+.......+\dfrac{3}{100.103}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+.......+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}.\dfrac{102}{103}=\dfrac{34}{103}\)
-
Andrey Trà My 10/03/2017 at 22:57\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}=\dfrac{1}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{1}{100.103}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}.\dfrac{102}{103}\)
\(=\dfrac{34}{103}\)
