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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
TB
4
6 tháng 8 2019
\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)
\(2S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{18.19.20}\)
\(=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\left(\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(=\frac{1}{2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{19.20}< \frac{1}{2}\)
\(2S< \frac{1}{2}\)
\(\Rightarrow S< \frac{1}{4}\) (ĐPCM)
CC
6 tháng 8 2019
\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{18.19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)\)
\(=\frac{1}{4}-\frac{1}{760}\)
=> S < \(\frac{1}{4}\)( vì 1/4 > 0)
NN
3
PV
22 tháng 4 2016
S = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/23.24.25
2.S = 2/1.2.3 + 2/2.3.4 + 2/3.4.5 + ... + 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 = 1/2 - 1/600
=> S = (1/2 - 1/600) : 2 = 1/4 - 1/1200
Dễ thấy S < 1/4 Hay S < 0,25
24 tháng 4 2016
1/2.(2/1.2.3+2/2.3.4+......2/23.24.25)
1/2.(1/1.2-1/2.3+1/2.3-1/3.4+……+1/23.24-1/24.25)
1/2.(1/1.2-1/24.25)
1/2.(1/2-1/600)
1/2.(300/600-1/600)
1/2.299/600
299/1200
Ta co 0.25=1/4
Nen ta so sanh 1/4 va 299/1200
Vi 300/1200>299/1200
Nen 1/4>299/1200
Ket luan 0,25>S
NL
1
17 tháng 5 2022
\(2S=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{23+24+25}=\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)+\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+...+\left(\dfrac{1}{23.24}-\dfrac{1}{24.25}\right)\)\(=\dfrac{1}{1.2}-\dfrac{1}{24.25}=\dfrac{299}{600}\)
Vậy \(S=\dfrac{299}{600}\div2=\dfrac{299}{1200}\)
24 tháng 2 2017
tui biết = mấy
kb và mỗi ngày k 3 nick nguyên huyên
sakura
Nguyễn Thị Thu Huyền
thì tui giải cho
NN
3
NT
17 tháng 5 2017
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+....+\dfrac{1}{2014.2015.2016}\\ =\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)+\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+.....+\left(\dfrac{1}{2014.2015}-\dfrac{1}{2015.2016}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2015.2016}\right)\\ =\dfrac{1}{4}-\dfrac{1}{2.2015.2016}< \dfrac{1}{4}\\ \Rightarrow A< \dfrac{1}{4}\)
17 tháng 5 2017
Giải:
Ta có: \(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2014.2015.2016}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{2014.2015.2016}\right)\)
\(=\dfrac{1}{2}\)\(\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+...+\dfrac{1}{2014.2015}-\dfrac{1}{2015.2016}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2015.2016}\right)\)
\(=\dfrac{1}{2}.\dfrac{1}{1.2}-\dfrac{1}{2}.\dfrac{1}{2015.2016}=\dfrac{1}{4}-\) \(\dfrac{1}{2.2015.2016}\)
Mà \(\dfrac{1}{2.2015.2016}>0\Leftrightarrow\dfrac{1}{4}-\dfrac{1}{2.2015.2016}< \dfrac{1}{4}\)
Vậy \(A< \dfrac{1}{4}\)
LT
1
17 tháng 4 2017
Ta có :
\(S=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+..............+\dfrac{1}{98.99.100}\)
\(S=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+................+\dfrac{2}{98.99.100}\right)\)
\(S=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...........+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)
\(S=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{99.100}\right)\)
\(S=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{9900}\right)\)
\(S=\dfrac{1}{2}.\dfrac{4949}{9900}\)
\(S=\dfrac{4949}{19800}\)
~ Chúc bn học tốt ~
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= \(1-\frac{1}{3^{2008}}\)
BT
0
\(S=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{98\cdot99\cdot100}\)
\(S=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\right)\)
\(S=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)
\(S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(S=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)
\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(\Rightarrow2S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(=\frac{1}{1.2}-\frac{1}{99.100}=\frac{4849}{9900}\)
\(\Rightarrow S=\frac{4949}{9900}\div2=\frac{4949}{19800}\)