a , Chứng tỏ rằng :
2/n(n+1)(n+2) =1/n(n+1)-1/(n+1)(n+2)
b, Áp dụng tính :
1/1.2.3+1/2.3.4+1/3.4.5+................+1/98.99.100
K
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24 tháng 9 2021
4A = 4.[1.2.3 + 2.3.4 + 3.4.5 + … + (n – 1).n.(n + 1)]
4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + … + (n – 1).n.(n + 1).4
4A = 1.2.3.4 + 2.3.4.(5 – 1) + 3.4.5.(6 – 2) + … + (n – 1).n.(n + 1).[(n + 2) – (n – 2)]
4A = 1.2.3.4 + 2.3.4.5 – 1.2.3.4 + 3.4.5.6 – 2.3.4.5 + … + (n – 1).n(n + 1).(n + 2) – (n – 2).(n – 1).n.(n + 1)
4A = (n – 1).n(n + 1).(n + 2)
A = (n – 1).n(n + 1).(n + 2) : 4.
21 tháng 8 2017
549 + X = 1326
X = 1326 - 549
X = 777
X - 636 = 5618
X = 5618 + 636
X = 6254
TH
13 tháng 2 2018
A = \(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)
3A= \(1+\frac{1}{3}+...+\frac{1}{3^{2006}}+\frac{1}{3^{2007}}\)
3A-A= \(1-\frac{1}{3^{2008}}\)
11 tháng 4 2018
1)
a)\(x+\dfrac{3}{22}=\dfrac{27}{121}.\dfrac{11}{9}\)
\(x+\dfrac{3}{22}=\dfrac{3}{11}\)
\(x=\dfrac{6}{22}-\dfrac{3}{22}\)
\(x=\dfrac{3}{22}\)
NP
0
PP
2
4 tháng 8 2016
\(N=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n.\left(n+1\right).\left(n+2\right)}\)
\(\Rightarrow2N=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n.\left(n+1\right).\left(n+2\right)}\)
\(\Rightarrow N=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n.\left(n+1\right)}-\frac{1}{\left(n+1\right).\left(n+2\right)}\right)\)
\(\Rightarrow N=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{\left(n+1\right).\left(n+2\right)}\right)\)
4 tháng 8 2016
N=1/1.2.3 +1/2.3.4 +1/3.4.5 +...+1/n.(n+1).(n+2)
⇒2N=2/1.2.3 +2/2.3.4 +2/3.4.5 +...+2/n.(n+1).(n+2)
⇒N=1/2 .(1/1.2 −1/2.3 +1/2.3 −1/3.4 +1/3.4 −1/4.5 +...+1/n.(n+1) −1/(n+1).(n+2) )
⇒N=1/2 .(1/1.2 −1/(n+1).(n+2) )
chúc bạn học tốt !
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