DH
1
K
Khách
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DH
1
VM
14 tháng 9 2017
Ta Có:
C=1/2+(1/2)^2+....+(1/2)^2017
=>2C=1+1/2+.......+(1/2)^2016
=>C=2C-C=1-(1/2)^2017=>C< hoặc =1(Dấu bằng xảy ra khi (1/2)^2017=0)
=>(đpcm)
KN
16 tháng 4 2019
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2017^2}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow A< 1-\frac{1}{2017}=\frac{2016}{2017}\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2017^2}< \frac{2016}{2017}\left(đpcm\right)\)
PL
2
20 tháng 9 2018
ta có: \(\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{2015^2}< \frac{1}{2014.2015};\frac{1}{2016^2}< \frac{1}{2015.1026};\frac{1}{2017^2}< \frac{1}{2016.2017}\)
=> 1/22 + 1/32 + 1/42 + ... + 1/20152 + 1/20162 + 1/20172 < 1/22 + (1/2.3 + 1/3.4 + ....+1/2014.2015 + 1/2015.2016 + 1/2016.2017)
= 1/4 + 1/2 - 1/2017 = 3/4- 1/2017 < 3/4
=> đ p c m
20 tháng 9 2018
ta có: \(\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{2015^2}< \frac{1}{2014.2015};\frac{1}{2016^2}< \frac{1}{2015.1026};\frac{1}{2017^2}< \frac{1}{2016.2017}\)
=> 1/22 + 1/32 + 1/42 + ... + 1/20152 + 1/20162 + 1/20172 < 1/22 + (1/2.3 + 1/3.4 + ....+1/2014.2015 + 1/2015.2016 + 1/2016.2017)
= 1/4 + 1/2 - 1/2017 = 3/4- 1/2017 < 3/4
=> đ p c m
12 tháng 12 2017
Ta có :
\(S=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{2016}+\left(\frac{1}{2}\right)^{2017}\)
\(2S=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2015}+\left(\frac{1}{2}\right)^{2016}\)
\(2S-S=\left[1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2015}+\left(\frac{1}{2}\right)^{2016}\right]-\left[\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{2016}+\left(\frac{1}{2}\right)^{2017}\right]\)
\(S=1-\left(\frac{1}{2}\right)^{2017}< 1\)
DN
0
NT
1
S
11 tháng 5 2017
\(C=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2017}{4^{2017}}\)
\(4C=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2017}{4^{2016}}\)
\(4C-C=\left(1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2017}{4^{2016}}\right)-\left(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2017}{4^{2017}}\right)\)
\(3C=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2016}}-\frac{2017}{4^{2017}}\)
\(12C=4+1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2015}}-\frac{2017}{4^{2016}}\)
\(12C-3C=\left(4+1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2015}}-\frac{2017}{4^{2016}}\right)-\left(1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2016}}-\frac{2017}{4^{2017}}\right)\)
\(9C=4-\frac{2017}{4^{2016}}-\frac{1}{4^{2016}}+\frac{2017}{4^{2017}}\)
\(9C=4-\frac{8068}{4^{2017}}-\frac{4}{4^{2017}}+\frac{2017}{4^{2017}}\)
\(9C=4-\frac{10081}{4^{2017}}\)
=> 9C < 4
=> C < \(\frac{4}{9}\)< \(\frac{1}{2}\)(đpcm)
Ta có:
C=1/2+(1/2)^2+....+(1/2)^2017
=> 2C=1+1/2+(1/2)^2+......+(1/2)^2016
=> C=2C-C=1-(1/2)^2017=>C< hoặc=1(Dấu bằng xảy ra khi (1/2)^2017=0)
=>(đpcm)