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\(=\dfrac{1}{3}.\left(\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\right)\)
\(=\dfrac{1}{3}.\left(\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\right)\)
\(=\dfrac{1353}{8120}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
s= (2/1.2.3 +2/2.3.4+...+2/98.99.100):2= (1/1.2-1/2.3+1/2.3-1/3.4+...+1/98.99-1/99.100):2=(1/1.2-1/99.100):2=4949/19800=>S=4949/19800
![](https://rs.olm.vn/images/avt/0.png?1311)
tao có:
2p=2/1.2.3+2/2.3.4+...+2/n.n(+1)n(n+2)
2p=3-1/1.2.3+4-2/1.2.3+...+(n+2)-n/n.(n+1).(n+2)
2p=3/1.2.3-1/1.2.3+4/2.3.4-2/2.3.4+...+(n+2)/n.(n+1).(n+2)-n/n.(n+1).(n+2)
2p=1/1.2-1/2.3+1/2.3-1/3.4+...+1/n.(n+1)-1/(n+1).(n+2)
2p=1/1.2-1/(n+1).(n+2)
2p=(n+!).(n+2)-2/(2n+2).(n+2)
suy ra p=(n+1).(n+2)-2/(2n+2).(2n+4)
2s=3-1/1.2.3+4-2/1.2.3+...+50-48/48.49.50
2s=3/1.2.3-1/1.2.3+4/2.3.4-2/2.3.4+...+50/49.50.48-48/48.50.49
2s=1/1.2-1/2.3+1/2.3-1/3.4+...+1/48.49-1/49.50
2s=1/1.2-1/49.50
'2s=1/2-1/2450
2s=1225/2450-1/2450
2s=1224/2450
s=612/1225
\(P=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)1
\(P=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\right)\)
\(P=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(P=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(P=\frac{\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)}{2}\)
S cx tinh giong v
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{8.9.10}\right).x=\dfrac{22}{45}\)
=> \(\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{8.9.10}\right).x=\dfrac{22}{45}\)
=> \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-...-\dfrac{1}{9.10}\right).x=\dfrac{22}{45}\)
=> \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{9.10}\right).x=\dfrac{22}{45}\)
=> \(\dfrac{1}{2}.\dfrac{22}{45}.x=\dfrac{22}{45}\)
=> \(\dfrac{1}{2}.x=1\)
=> \(x=2\)
Vậy x = 2
Chúc bạn học tốt !!!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2009.2010.2011}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2009.2010}-\frac{1}{2010.2011}\)
\(=\frac{1}{2}-\frac{1}{2010.2011}< \frac{1}{2}\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1/1.2.3 + 1/2.3.4 + ... + 1/23.24.25
2A=2/1.2.3 + 2/2.3.4 + ... + 2/23.24.25
=1/1.2 - 1/2.3 + 1/2.3 -1/3.4 + .... + 1/23.24 - 1/24.25
=1/1.2 - 1/24.25
Tớ chỉ giải đến đó thôi còn lại các bạn cứ bấm máy tính là ra
Bài toán trên áp dụng bài toán tổng quát sau:
\(\frac{1}{n.\left(n+1\right)}-\frac{1}{\left(n+1\right).\left(n+2\right)}=\frac{2}{n.\left(n+1\right).\left(n+2\right)}\)
Suy ra
\(\frac{1}{n.\left(n+1\right).\left(n+2\right)}=\left(\frac{1}{n.\left(n+1\right)}-\frac{1}{\left(n+1\right).\left(n+2\right)}\right).\frac{1}{2}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{23.24.25}\)
\(=\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{23.24}-\frac{1}{24.25}\right).\frac{1}{2}\)
\(=\left(\frac{1}{1.2}-\frac{1}{24.25}\right).\frac{1}{2}\)
\(=\frac{299}{1200}\)