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B = ( -1 ) - 2 + ( - 3 ) - 4 + ... + ( - 49 ) - 50 Có 50 số hạng
B = ( - 3 ) +( - 7 ) + .... + ( - 99 ) có 50 : 2 = 25 số hạng
Tổng B là [( - 99 ) + ( - 3 ) ] x 25 : 2 = ( - 1275 )
Vậy B = -1275
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(A=\frac{1}{2}.\frac{4949}{9900}\)
\(A=\frac{4949}{19800}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
B = ( -1 ) + 2 + ( -3 ) + 4 + ... + ( -49 ) + 50
=> B = [ 2 + ( -1 ) ] + [ ( -3 ) + 4 ] + ... + [ ( -49 ) + 50 ] ( 25 cặp số )
=> B = 1 + 1 + ... + 1
=> B = 1 x 25
=> B = 25
Vậy B = 25
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+....+\frac{1}{1+2+3+..+50}\)
\(=\frac{1}{\frac{2\left(2+1\right)}{2}}+\frac{1}{\frac{3\left(3+1\right)}{2}}+\frac{1}{\frac{4\left(4+1\right)}{2}}+....+\frac{1}{\frac{2015\left(2015+1\right)}{2}}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{2015.2016}\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2016}\right)=\frac{1007}{1008}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
- 1 - 2 - 3 - ... - 49 - 50
= - (1 + 2 + 3 + ... + 49 + 50)
= - [(1 + 50) . 50 : 2]
= - [51 . 50 : 2]
= - [2550 : 2]
= - 1275
G = \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{49.50}\)
G = \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
G = \(\dfrac{1}{1}\) - \(\dfrac{1}{50}\)
G = \(\dfrac{49}{50}\)