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số số hạng là : ( 99.100 - 1.2 ) / 1.1 +1=90 ( số )
tổng của S là : ( 1.2 + 99.100 ) * 90 / 2=4513.5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3A=1.2.3+2.3.\left(4-1\right)+...+100.101.\left(102-99\right)\)
\(3A=1.2.3+2.3.4-1.2.3+.......+100.101.102-99.100.101\)
\(3A=100.101.102\)
\(A=\frac{100.101.102}{3}\)
\(A=343400\)
3=1.2.3+2.3(4-1)+...+100.101(102-99)
3=1.2.3+2.3.4-1.2.3+.....+100.101.102-99.100.101
3=100.101.101
=100.101.102/3
=343400
mn ủng hộ ^--^
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi A là biểu thức ta có:
A = 1.2+2.3+3.4+......+99.100
Gấp A lên 3 lần ta có:
A . 3 = 1.2.3 + 2.3.3 + 3.4.3 + … + 99.100.3
A . 3 = 1.2.3 + 2.3.(4 - 1) + 3.4.( 5 - 2) + … + 99.100. (101 - 98)
A . 3 = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + … + 99.100.101 - 98.99.100
A . 3 = 99.100.101
A = 99.100.101 : 3
A = 33.100.101
A = 333 300
A= 1.2+2.3+3.4+...+99.100
3A= 1.2.3+2.3.4+3.4.3+...+99.100.3
3A= 1.2.(3-0)+2.3(4-1)+3.4(5-2)+...+ 99.100(101-98)
3A= (1.2.3+2.3.4+3.4.5+...+ 99.100.110)-(0.1.2+1.2.3+2.3.4+3.4.5+...+98.99.100)
3A= 99.100.101- 0.1.2
3A= 999 900-0
3A= 999 900
A= 999 900:3
A= 333 300
GOOD LUCK !!!! ^ ^
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
đặt A = 1.2 + 2.3 + 3.4 + ... + 99.100
=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3
=> 3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98)
=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100
=> 3A = 99.100.101
=> A = 99.100.101 : 3
=> A = 333300
Tính tổng dãy sau :
Bài giải :
Đặt S = 1 . 2 + 2 . 3 + 3 . 4 + .... + 99 . 100
3S = 1 . 2 . 3 + 2 . 3 . 3 + 3 . 4 . 3 + ... + 98 . 99 . 3 + 99 . 100 . 3
= 1 . 2 . 3 + 2 . 3 ( 4 - 1 ) + 3 . 4 ( 5 - 2 ) + ... + 98 . 99 ( 100 - 97 ) + 99 . 100 ( 101 - 98 )
= 1 . 2 . 3 + 2 . 3 . 4 - 1 . 2 . 3 + 3 . 4 . 5 - 2 . 3 . 4 + ... - 97 . 98 . 99 + 99 . 100 . 101 - 98 . 99 . 100
3S = 99 . 100 .101
=> S = 99 . 100 .101 : 3
= ( 99 : 3 ) . ( 100 . 101 )
= 33 . 10 100
= 333 300
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1/1.2+1/2.3+1/3.4+...+1/2005.2006=(1-1/2)+(1/2-1/3)+...+(1/2005-1/2006)=1-1/2+1/2-1/3+...+1/2005-1/2006
=1-(1/2-1/2)+...-1/(1/2005-1/2005)-1/2006=1-1/2006=2005/2006
k mình nha
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B = 1.2+2.3+3.4+...+99.100
B=1.100
B=100
C=1.3+2.4+3.5+4.6+...+9.11
C=1.(2+1)+2.(3+1)+3.(4+1)+4.(5+1)+...+9.(10+1)
C=1.2+1+2.3+1+3.4+1+4.5+1+...+9.10+1
C=(1.2+2.3+3.3+4.5+...+9.10)+(1+1+1+1+..+1)
C=1.10+10
C=10+10
C=20
a) B = 1.2+2.3+3.4+..+99.100
=>3B=1.2.3+2.3.3+3.4.3+...+99.100.3
3B = 1.2.3+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)
3B = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5-2.3.4+...+99.100.101-98.99.100
3B = (1.2.3+2.3.4+3.4.5+..+99.100.101) - (1.2.3+2.3.4+...+98.99.100)
3B = 99.100.101
\(B=\frac{99.100.101}{3}=333300\)
b) C = 1.3+2.4+3.5+4.6+...+9.11
C = (2-1).(2+1)+(3-1).(3+1) + (4-1).(4+1)+(5-1).(5+1)+...+(10-1).(10+1)
C = 22 - 1 + 32 - 1 + 42 - 1 + 52 - 1 +...+102 - 1
C = (22+32+42+52+...+102) -(1+1+...+1)
...
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Ta có: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2020\cdot2021}+\dfrac{1}{2021\cdot2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
1/1x2+1/2x3+1/3x4+...+1/2020x2021+1/2021x2022
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2020-1/2021+1/2021-1/2022.
=1/1-1/2022
=2021/2022
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=9,8+8,7+7,6+...+2,1-1,2-2,3-3,4-...-8,9\)
\(=\left(9,8-8,9\right)+\left(8,7-7,8\right)+...+\left(2,1-1,2\right)\)
\(=0,9+0,9+...+0,9\)
\(=0,9\times8=7,2\)
1.2 + 2.3 + 3.4 + ....... + 30.31
= \(\frac{30.31.32}{3}\) = 9920
1*2+2*3+3*4+...+30*31=2660