NT
0
K
Khách
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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:
\(2C=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{a\left(a+1\right)\left(a+2\right)}\)
\(2C=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}\)
\(2C=\frac{1}{1.2}-\frac{1}{\left(a+1\right)\left(a+2\right)}\)
\(C=\left[\frac{1}{2}-\frac{1}{\left(a+1\right)\left(a+2\right)}\right]:2=\frac{\left(a+1\right)\left(a+2\right)-2}{4\left(a+1\right)\left(a+2\right)}=\frac{a\left(a+3\right)}{4\left(a+1\right)\left(a+2\right)}\)
TP
2
25 tháng 7 2016
Câu a)
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(=\left(2^{100}+2^{99}+2^{98}+2^{97}+...+2^2+2\right)-2\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)\)
\(=\left(2^{100}+2^{99}+2^{98}+2^{97}+...+2^2+2\right)-\left(2^{100}+2^{98}+2^{96}+...+2^4+2^2\right)\)
\(=2^{99}+2^{97}+2^{95}+...+2^3+2\)
\(=\frac{2^2\cdot\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)-\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)}{3}\)
\(=\frac{\left(2^{101}+2^{99}+2^{97}+...+2^5+2^3\right)-\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)}{3}\)
\(=\frac{2^{101}-2}{3}\)
6 tháng 4 2017
\(2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2015.2016.2017}\)
\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{2.4}+...+\frac{1}{2015.2016}-\frac{1}{2016.2017}\)
\(2B=\frac{1}{1.2}-\frac{1}{2016.2017}\)
\(B=\frac{\frac{1}{1.2}-\frac{1}{2016.1017}}{2}\)
D
3
14 tháng 3 2017
=1+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{2}\) -\(\frac{1}{3}\) -\(\frac{1}{4}\)+\(\frac{1}{3}\) - \(\frac{1}{4}\)-\(\frac{1}{5}\)+.....+\(\frac{1}{99}\)-\(\frac{1}{100}\)-\(\frac{1}{101}\)
=1+\(\frac{1}{101}\)
=\(\frac{102}{101}\)
14 tháng 3 2017
1/1.2.3 = 1/2 .[1/1.2 - 1 / 2.3]
1/2.3.4 = 1/2[ 1/2- 1/3 ]
...................
1/99.100.101 = 1/2[ 1/99. 100 - 1/100.101]
=> A= 1/2 [ 1/1.2- 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/ 4.5 +.........+ 1/99 .100 - 1/100. 101]
A = 1/2 . [1/1.2 -1/100 .101]
A= 1/2 . 5049 /10100 = 5049 / 20200.
Mình nghĩ là vậy đó.
NB
0
NT
3
HT
2 tháng 9 2015
A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+..+\frac{1}{99.100.101}\)
A = \(\frac{1}{2}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{101-99}{99.100.101}\right)\)
A = \(\frac{1}{2}.\left(\frac{3}{1.2.3}-\frac{1}{1.2.3}+\frac{4}{2.3.4}-\frac{2}{2.3.4}+...+\frac{101}{99.100.101}-\frac{99}{99.100.101}\right)\)
A = \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
A = \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{100.101}\right)\)
A = \(\frac{1}{2}.\frac{5049}{10100}\)
A = \(\frac{5049}{20200}\)
24 tháng 5 2015
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2008}-\frac{1}{2009}=\frac{1}{1}+\left(-\frac{1}{2}+\frac{1}{2}\right)+...+\left(-\frac{1}{2008}+\frac{1}{2008}\right)-\frac{1}{2009}\)
\(A=1-\frac{1}{2009}=\frac{2008}{2009}\)
\(2.B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2008.2009.2010}\)
\(2.B=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\left(\frac{1}{2008.2009}-\frac{1}{2009.2010}\right)\)
\(2.B=\frac{1}{1.2}+\left(-\frac{1}{2.3}+\frac{1}{2.3}\right)+...+\left(-\frac{1}{2008.2009}+\frac{1}{2008.2009}\right)-\frac{1}{2009.2010}\)
\(2.B=\frac{1}{1.2}-\frac{1}{2009.2010}=\frac{2009.2010-1.2}{2009.2010}\)
=> \(B=\frac{2009.1005-1}{2009.2010}\)
Vậy \(\frac{B}{A}=\frac{2009.1005-1}{2009.2010}:\frac{2008}{2009}=\frac{2009.1005-1}{2008.2010}=...\)
NT
0
LH
1
PV
12 tháng 4 2016
Đặt S = 1/1.2.3 - 1/2.3.4 - 1/3.4.5 - ...- 1/97.98.99
S x 2 = 2/1.2.3 - 2/2.3.4 - 2/3.4.5 - ...- 2/97.98.99
= (1/1.2 -1/2.3) - (1/2.3 - 1/3.4 ) - (1/3.4 - 1/4.5) - ...- (1/97.98 - 1/98.99)
= 1/1.2 - 1/2.3 - 1/2.3 + 1/3.4 - 1/3.4 + 1/4.5 - ....- 1/97.98 + 1/98.99
= 1/2 -1/3 + 1/98.99
= 1618/9072 => S = 1618/9072 : 2 = 809/9072