C = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 +...+ 1/2008.2009.2010
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LB
7 tháng 5 2018
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
8 tháng 5 2018
\(P=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)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
DT
0
ND
1
13 tháng 11 2015
\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
=> \(2C=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{98.99}-\frac{1}{99.100}\)
\(=\frac{1}{2}-\vec{\frac{1}{99.100}=\frac{4949}{99.100}}\)
\(C=\frac{4949}{2.99.100}\)
S
8 tháng 10 2017
Từ công thức \(\frac{2}{a\left(a+1\right)\left(a+2\right)}=\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}\), ta có:
\(2C=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{a\left(a+1\right)\left(a+2\right)}\)
\(2C=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}\)
\(2C=\frac{1}{1.2}-\frac{1}{\left(a+1\right)\left(a+2\right)}\)
\(C=\left[\frac{1}{2}-\frac{1}{\left(a+1\right)\left(a+2\right)}\right]:2=\frac{\left(a+1\right)\left(a+2\right)-2}{4\left(a+1\right)\left(a+2\right)}=\frac{a\left(a+3\right)}{4\left(a+1\right)\left(a+2\right)}\)
23 tháng 1 2022
2P=2/1.2.3+2/2.3.4+2/3.4.5+2/10.11.12
2P=1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+.....+1/10.11-1/11.12
2P=1/1.2-1/11.12
2P=1/2-1/132
2P=66/132-1/132
2P=65/132
P=65/264
23 tháng 1 2022
\(P=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{10.11.12}\)
\(P=\dfrac{1}{2}-\dfrac{1}{11.12}\)
\(P=\dfrac{65}{132}\)
29 tháng 11 2016
Đặt A = 1.2.3 + 2.3.4 + 3.4.5 + ... + 28.29.30
4A = 1.2.3.(4-0) + 2.3.4.(5-1) + 3.4.5.(6-2) + ... + 28.29.30.(31-27)
4A = 1.2.3.4 - 0.1.2.3. + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 28.29.30.31 - 27.28.29.30
4A = 28.29.30.31 - 0.1.2.3
4A = 28.29.30.31
\(A=\frac{28.29.30.31}{4}=7.29.30.31=188790\)
Theo cách tính trên ta dễ dàng tính được:
1.2.3 + 2.3.4 + 3.4.5 + ... + (n - 1).n.(n + 1) = \(\frac{\left(n-1\right).n.\left(n+1\right).\left(n+2\right)}{4}\)
17 tháng 3 2017
= 1 . 1/2 . 1/3 + 1/2 . 1/3 . 1/4 + ... + 1/37 . 1/38 . 1/39
= 1 . 1/39
= 1/39
Mong moi nguoi chi them03
PT
0
HT
2 tháng 8 2015
1/1.2.3 + 1/2.3.4 +....+1/98.99.100
= 1/2 . (3-1/1.2.3 + 4-2/2.3.4 +....+ 100-98/98.99.100)
= 1/2 . (3/1.2.3 -1/1.2.3 + 4/2.3.4 - 2/2.3.4 +.......+ 100/98.99.100 - 98/98.99.100)
= 1/2 . (1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 +......+ 1/98.99 - 1/99.100)
= 1/2 . (1/2 - 1/9900)
= 1/2 . 4949/9900
= 4949/19800