1x2+3x4+5x6+.............................+99x100
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VT
3
15 tháng 3 2016
Gọi biểu thức trên là A, ta có :
A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
A x 3 = 99x100x101
A = 99x100x101 : 3
A = 333300
TH
12
10 tháng 6 2017
A = 2 + ( 2+ 1).4 + ( 4 + 1)6 + … + (98 + 1).100
= 2 + 2.4 + 4 + 4.6 + 6 + … + 98.100 + 100
= (2.4 + 4.6 + … + 98.100 ) + (2 + 4 + 6 + … + 100)
= 98.100.102 : 6 + 102.50:2
= 166600 + 2550
= 169150
F
0
PM
2
11 tháng 9 2015
A = 1 x 2 + 2 x 3 + ....... + 10 x 11
3A = 1 x 2 x 3 + 2 x 3 x 3 + ..........+ 10 x 11 x 3
3A = 1 x 2 x (3-0) + 2 x 3 x (4-1) + .......... + 10 x 11 x (12 -9)
3A = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + ........... + 10 x 11 x 12 - 9 x 10 x 11
3A = (1 x 2 x 3 - 1 x 2 x 3) + ( 2 x 3 x 4 - 2 x 3 x 4) +............ + 10 x 11 x 12
3A = 10 x 11 x 12 = 1320
A = 1320 : 3 = 440
V
7 tháng 7 2019
Gọi biểu thức trên là A, ta có :
A= 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
A x 3 = 99x100x101 A = 99x100x101 : 3 A = 333300
V
1
10 tháng 6 2017
B x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
= 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
= 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
= 99x100x101
B = 99x100x101 : 3
= 333300
nhanh k minh
XR
10 tháng 6 2017
B= 1x2+3x4+5x6+...+99x100
=> Bx3= 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ...+ 99x100x3
=> Bx3= 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3)+...+99x100x(101-98)
=> Bx3= 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 +4x5x6 - 3x4x5 +...+ 99x100x101 - 98x99x100
=> Bx3= 99x100x101
=> B= 99x100x101:3
=> B= 333300
BN
15 tháng 7 2017
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{100}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{100}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)
\(=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
\(=\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}\right)+\left(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}\right)\)
Ta có:
\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}>\dfrac{1}{75}+\dfrac{1}{75}+...+\dfrac{1}{75}=\dfrac{25}{75}=\dfrac{1}{3}\)
\(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}>\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{25}{100}=\dfrac{1}{4}\)
\(\Rightarrow A>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}\) (1)
Lại có:
\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{25}{50}=\dfrac{1}{2}\)
\(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}< \dfrac{1}{75}+\dfrac{1}{75}+...+\dfrac{1}{75}=\dfrac{25}{75}=\dfrac{1}{3}\)
\(\Rightarrow A< \dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\) (2)
Từ (1) và (2) suy ra \(\dfrac{7}{12}< A< \dfrac{5}{6}\)
PM
Cho A=\(\frac{1}{1x2}+\frac{1}{3x4}+\frac{1}{5x6}+....+\frac{1}{99x100}\)
Chứng minh rằng: 7/12<A<5/6
4
AM
1 tháng 7 2015
\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
Do \(\frac{1}{51}>\frac{1}{52}>...>\frac{1}{100}\Rightarrow A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}>25\cdot\frac{1}{80}+25\cdot\frac{1}{100}=\frac{7}{12}\)
và \(A