Tính:A=1/98.99.100+1/99.100.101+1/100.101.102+1/101.102.103
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NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
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/
\(\dfrac{2}{5}.d=\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}+\dfrac{101-99}{99.100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}+\dfrac{1}{99.100}-\dfrac{1}{100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{100.101}\Rightarrow d=\left(\dfrac{1}{2.3}-\dfrac{1}{100.101}\right):\dfrac{2}{5}\)
PT
4
14 tháng 4 2019
A=1/2 *(1/1*2-1/2*3+1/2*3-1/3*4+........+1/98*99-1/99*100)
=1/2*(1/2-1/99*100)
=1/2*(4950-1/9900)
=4950/19800
14 tháng 4 2019
\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)
\(A=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\right]\)
\(A=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+....+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right]\)
\(A=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{99\cdot100}\right]=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)
TT
1
AH
Akai Haruma
Giáo viên
3 tháng 11 2023
Lời giải:
Gọi tổng trên là A
$2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{101.102.103}$
$=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{103-101}{101.102.103}$
$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{101.102}-\frac{1}{102.103}$
$=\frac{1}{1.2}-\frac{1}{102.103}=\frac{2626}{5253}$
$\Rightarrow A=\frac{1313}{5253}$
TT
1
AH
Akai Haruma
Giáo viên
3 tháng 11 2023
Lời giải:
Gọi tổng trên là A
$2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{101.102.103}$
$=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{103-101}{101.102.103}$
$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{101.102}-\frac{1}{102.103}$
$=\frac{1}{1.2}-\frac{1}{102.103}=\frac{2626}{5253}$
$\Rightarrow A=\frac{1313}{5253}$
DM
1
20 tháng 4 2016
Ta có: M=\(\frac{1}{1.2.3}\) +\(\frac{1}{2.3.4}\) +\(\frac{1}{3.4.5}\) +...+\(\frac{1}{100.101.102}\)
M=2.(\(\frac{1}{1.2.3}\) +\(\frac{1}{2.3.4}\) +\(\frac{1}{3.4.5}\) +...+\(\frac{1}{100.101.102}\) ).\(\frac{1}{2}\)
M=(\(\frac{2}{1.2.3}\) +\(\frac{2}{2.3.4}\) +\(\frac{2}{3.4.5}\) +...+\(\frac{2}{100.101.102}\) ).\(\frac{1}{2}\)
M=(\(\frac{1}{1.2}\) -\(\frac{1}{2.3}\) +\(\frac{1}{2.3}\) -\(\frac{1}{3.4}\) +\(\frac{1}{3.4}\) -\(\frac{1}{4.5}+...+\frac{1}{100.101}-\frac{1}{101.102}\) ).\(\frac{1}{2}\)
M=( \(\frac{1}{1.2}-\frac{1}{101.102}\)).\(\frac{1}{2}\)
Mà \(\frac{1}{1.2}-\frac{1}{101.102}<1\)
Và \(\frac{1}{2}<1\)
\(=>\) (\(\frac{1}{1.2}-\frac{1}{101.102}\) ) .\(\frac{1}{2}\) \(<1\)
\(=>\) M <1
10 tháng 5 2016
\(S=\frac{5}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(S=\frac{5}{2}.\left(\frac{1}{2.3}-\frac{1}{100.101}\right)\)
\(S=\frac{5}{2}.\left(\frac{5047}{30300}\right)\Rightarrow S=\frac{5047}{12120}\)
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
Ta có : \(A=\frac{1}{98.99.100}+\frac{1}{99.100.101}+\frac{1}{100.101.102}+\frac{1}{101.102.103}\)
\(2A=2.\left(\frac{1}{98.99.100}+\frac{1}{99.100.101}+\frac{1}{100.101.102}+\frac{1}{101.102.103}\right)\)
\(2A=\frac{2}{98.99.100}+\frac{2}{99.100.101}+\frac{2}{100.101.102}+\frac{2}{101.102.103}\)
\(2A=\frac{1}{98.99}-\frac{1}{99.100}+\frac{1}{99.100}-\frac{1}{100.101}+\frac{1}{100.101}-\frac{1}{101.102}+\frac{1}{101.102}-\frac{1}{102.103}\)
\(2A=\frac{1}{98.99}-\frac{1}{102.103}\)
\(2A=\frac{1}{9702}-\frac{1}{10506}\)
\(A=\frac{\left(\frac{1}{9702}-\frac{1}{10506}\right)}{2}\)
\(A=\left(\frac{1}{9702}-\frac{1}{10506}\right).\frac{1}{2}\)
\(A=\frac{1}{9702.2}-\frac{1}{10506.2}\)
\(A=\frac{1}{19404}-\frac{1}{21012}\)
\(A=\frac{21012-19404}{19404.21012}\)
\(A=\frac{1608}{19404.21012}\)
\(A=\frac{134}{19404.1751}\)
\(A=\frac{67}{9702.1751}\)