1.2+2.3+3.4+...97.98
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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S = 1.2 + 2.3 + 3.4 + ... + 97.98
=> 3S = 1.2.3 + 2.3.3 + 3.4.3 + ... + 97.98.3
= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 97.98.(99 - 96)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 97.98.99 - 96.97.98
= 97.98.99
= 941 094
=> S = 941 094 : 3 = 313698
Vậy S = 313698
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A = 1.2 + 2.3 + 3.4 + ... + 99.100
3A = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + ... + 99.100.(101-98)
3A = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100
3A = 99.100.101
A = 33.100.101
A = 333300
\(A=1.2+2.3+3.4+4.5+...+97.98+98.99+99.100\)
\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+4.5.\left(6-3\right)+...+99.100.\left(101-98\right)\)
\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100\)
\(3A=99.100.101\)
\(A=\frac{99.100.101}{3}=\frac{999900}{3}=333300\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 1 - \(\dfrac{1}{1.2}\) - \(\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}...-\dfrac{1}{97.98}\)
A= 1-\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{97.98}\right)\)
A= 1- \(\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}...+\dfrac{1}{97}-\dfrac{1}{98}\right)\)
A= 1- \(\left(\dfrac{1}{1}-\dfrac{1}{98}\right)\)
A=1- 1 + \(\dfrac{1}{98}\)
A= \(\dfrac{1}{98}\)
Lời giải:
$1-A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{97.98}$
$1-A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{98-97}{97.98}$
$1-A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}$
$=1-\frac{1}{98}$
$\Rightarrow A=\frac{1}{98}$
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt tổng trên = A
Có : 3A = 1.2.3+2.3.3+....+98.99.3
= 1.2.3+2.3.(4-1)+.....+98.99.(100-97)
= 1.2.3+2.3.4-1.2.3+.....+98.99.100-97.98.99
= 98.99.100
=> A = 98.99.100/3 = 323400
k mk nha
Gọi A = 1.2 + 2.3 + .. + 98.99
3A = 1.2.3 + 2.3.3 + ... + 98.99.3
3A = 1.2.3 + 2.3.(4 - 1) + ... + 98.99.(100 - 97)
3A = 1.2.3 + 2.3.4 - 1.2.3 + ... + 98.99.100 - 97.98.99
3A = 98.99.100
3A = 970200
A = 323400
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}\)
\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{99-98}{98.99}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}\)
\(=1-\dfrac{1}{99}\)
\(A=\dfrac{2021}{2022}=\dfrac{2022-1}{2022}=1-\dfrac{1}{2022}\)
Có \(2022>99>0\Leftrightarrow\dfrac{1}{99}>\dfrac{1}{2022}\)
Suy ra \(A>B\).
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A=1.98+2.97+3.96+...+96.3+97.2+98.1
B=1.2+2,3+3.4+...+96.97+97.98+98.99
Ta có: A=1+(1+2)+...+(1+2+3+...+97+98)
=\(\dfrac{1.2}{2}+\dfrac{2.3}{2}+...+\dfrac{98.99}{3}\)
=\(\dfrac{1.2+2.3+3.4+4.5+...+98.99}{2}\)=\(\dfrac{B}{2}\)
=>E=\(\dfrac{B}{2}\):2=\(\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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