1x4+4x7+7x10+...+2022x2023
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PP
2
NH
6 tháng 6 2019
\(\frac{11}{1.4}+\frac{11}{4.7}+...+\frac{11}{100.103}\)
\(=\frac{11}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(=\frac{11}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=\frac{11}{3}\left(1-\frac{1}{103}\right)\)
Tự tính
XO
6 tháng 6 2019
\(\frac{11}{1.4}+\frac{11}{4.7}+...+\frac{11}{100.103}\)
= \(\frac{11}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
= \(\frac{11}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
= \(\frac{11}{3}.\left(1-\frac{1}{103}\right)\)
= \(\frac{11}{3}.\frac{102}{103}\)
= \(\frac{374}{103}\)
BH
0
ND
1
HQ
Hà Quang Minh
Giáo viên
1 tháng 9 2023
\(A=\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+...+\dfrac{1}{37\times40}\\ =\dfrac{1}{3}\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+...+\dfrac{3}{37\times40}\right)\\ =\dfrac{1}{3}\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{40}\right)\\ =\dfrac{1}{3}\times\left(1-\dfrac{1}{40}\right)\\ =\dfrac{1}{3}\times\dfrac{39}{40}\\ =\dfrac{13}{40}\)
17 tháng 12 2022
\(=\dfrac{2}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{31\cdot34}\right)\)
\(=\dfrac{2}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)\)
\(=\dfrac{2}{3}\cdot\dfrac{33}{34}=\dfrac{11}{17}\)
VN
Võ Ngọc Phương
VIP
8 tháng 8 2023
a) \(\dfrac{2}{1\times4}+\dfrac{2}{4\times7}+\dfrac{2}{7\times10}+...+\dfrac{2}{97\times100}\)
\(=2.\left(\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+...+\dfrac{1}{97\times100}\right)\)
\(=2.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=2.\left(1-\dfrac{1}{100}\right)\)
\(=2.\dfrac{99}{100}\)
\(=\dfrac{99}{50}\)
_____
b) \(\dfrac{3}{1\times5}+\dfrac{3}{5\times9}+\dfrac{3}{9\times13}+...+\dfrac{3}{97\times101}\)
\(=3.\left(\dfrac{1}{1\times5}+\dfrac{1}{5\times9}+\dfrac{1}{9\times13}+...+\dfrac{1}{97\times101}\right)\)
\(=3.\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)
\(=3.\left(1-\dfrac{1}{101}\right)\)
\(=3.\dfrac{100}{101}\)
\(=\dfrac{300}{101}\)
30 tháng 6 2017
Đặt \(B=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+......+\frac{2}{100\cdot103}\)
\(B=\frac{2}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=\frac{2}{3}\cdot\left(1-\frac{1}{103}\right)\)
\(B=\frac{2}{3}\cdot\frac{102}{103}\)
\(\Rightarrow B=\frac{68}{103}\)
ML
30 tháng 6 2017
Đặt \(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{100.103}\)
\(A=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(A=\frac{2}{3}\left(1-\frac{1}{103}\right)\)
\(A=\frac{2}{3}\cdot\frac{102}{103}\)
\(A=\frac{68}{103}\)
VH
4