5/1.2.3+5/2.3.4+....+5/99.100.101
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Nguyễn Ngọc Anh Minh
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20 tháng 7 2023
a/
\(b=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)
\(2b=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{99-97}{97.99}=\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}=\)
\(=1-\dfrac{1}{99}=\dfrac{98}{99}\Rightarrow b=\dfrac{98}{2.99}=\dfrac{49}{99}\)
b/
\(c=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}=\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+\dfrac{1}{98.99}-\dfrac{1}{99.100}=\)
\(=\dfrac{1}{2}-\dfrac{1}{99.100}\)
c/
\(\dfrac{2}{5}.d=\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}+\dfrac{101-99}{99.100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}+\dfrac{1}{99.100}-\dfrac{1}{100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{100.101}\Rightarrow d=\left(\dfrac{1}{2.3}-\dfrac{1}{100.101}\right):\dfrac{2}{5}\)
16 tháng 8 2018
A = 1.2.3 + 2.3.4 + ... + 99.100.101
=> 4A = 1.2.3.4 + 2.3.4.4 + ... + 99.100.101.4
=> 4A = 1.2.3.4 + 2.3.4.(5 - 1) + ... + 99.100.101.(102 - 98)
=> 4A - 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + ... + 99.100.101.102 - 98.99.100.101
=> 4A = 99.100.101.102
=> A = 99.100.101.102 :4
=> A = 25497450
18 tháng 7 2017
Đặt \(A=1.2.3+2.3.4+3.4.5+...+99.100.101\)
\(\Rightarrow4A=1.2.3.4+2.3.4.4+...+99.100.101.4\)
\(=1.2.3\left(4-0\right)+2.3.4\left(5-1\right)+...+99.100.101\left(102-98\right)\)
\(=\left(1.2.3.4+2.3.4.5+...+99.100+101.102\right)-\left(0.1.2.3+1.2.3.4+...+98.99.100.101\right)\)
\(=99.100.101.102-0.1.2.3\)
\(=101989800\)
\(\Rightarrow A=101989800:4=25497450\)
Vậy \(A=25497450.\)
13 tháng 9 2018
Đặt A = 1.2.3 + 2.3.4 + ... + 99.100.101
=> 4A = 1.2.3.4 + 2.3.4.(5-1) + ... + 99.100.101.(102-98)
=> 4A = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + ... + 99.100.101.102 - 98.99.100.101
=> 4A = 99.100.101.102
=> 4A = 101989800
=> A = 25497450
TV
0
17 tháng 10 2017
1.2.3 = 1/4 . (1.2.3.4 - 0.1.2.3)
2.3.4 = 1/4 . (2.3.4.5 - 1.2.3.4)
3.4.5 = 1/4 . (3.4.5.6 - 2.3.4.5)
.................
99.100.101 = 1/4 . (99.100.101.102 - 98.99.100.101)
C = 1.2.3+2.3.4+3.4.5+.........+99.100.101
C= 1/4 . (99.100.101.102 - 98.99.100.101)
CHUC BN HOK GIỎI!
10 tháng 2 2017
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10100}\right)=\frac{5049}{20200}\)
10 tháng 2 2017
A=1/2(1/1.2-1/2.3+1/2.3-1/3.4+.......+1/99.100-1/100.101)
A=1/2(1/1.2-1/100.101)
A=1/2(1/1.2-1/100.101)=5049/202000
\(\dfrac{5}{1\cdot2\cdot3}+\dfrac{5}{2\cdot3\cdot4}+...+\dfrac{5}{99\cdot100\cdot101}\)
\(=5\cdot\left(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{99\cdot100\cdot101}\right)\)
\(=5\cdot\dfrac{1}{2}\cdot\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}-\dfrac{1}{100\cdot101}\right)\)
\(=\dfrac{5}{2}\cdot\left(\dfrac{1}{1\cdot2}-\dfrac{1}{100\cdot101}\right)\)
\(=\dfrac{5}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{10100}\right)\)
\(=\dfrac{5}{2}\cdot\dfrac{5049}{10100}\)
\(=\dfrac{25245}{20200}=\dfrac{5049}{4040}\)