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16 tháng 2 2020
Ta có:
\(A=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{100}}\)
\(\Rightarrow2^2A=1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}\)
\(\Rightarrow4A=1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}\)
\(\Rightarrow4A-A=1-\frac{1}{2^{100}}< 1\Rightarrow3A< 1\Rightarrow A< \frac{1}{3}\left(đpcm\right)\)
MN
5 tháng 1
A=221+241+261+...+21001
\(\Rightarrow 2^{2} A = 1 + \frac{1}{2^{2}} + \frac{1}{2^{4}} + . . . + \frac{1}{2^{98}}\)
\(\Rightarrow 4 A = 1 + \frac{1}{2^{2}} + \frac{1}{2^{4}} + . . . + \frac{1}{2^{98}}\)
\(\Rightarrow 4 A - A = 1 - \frac{1}{2^{100}} < 1 \Rightarrow 3 A < 1 \Rightarrow A < \frac{1}{3} \left(\right. đ p c m \left.\right)\)
TT
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HV
0
ZS
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LO
1
23 tháng 12 2018
\(2^2A=1+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(4A-A=1-\frac{1}{2^{100}}\)
\(A=\frac{1-\frac{1}{2^{100}}}{3}\)
4 tháng 6 2019
a) \(\left(\frac{-1}{3}\right)^4=\frac{\left(-1\right)^4}{3^4}=\frac{1}{81}\)
b) \(\left(-2\frac{1}{4}\right)^3=\left(\frac{-9}{4}\right)^3=\frac{\left(-9\right)^3}{4^3}=\frac{-729}{64}\)
c) \(\left(-0,2\right)^2=\left(\frac{-1}{5}\right)^2=\frac{\left(-1\right)^2}{5^2}=\frac{1}{25}\)
d) \(\left(-5,3\right)^0=1\)
4 tháng 6 2019
a)\(\left(\frac{-1}{3}\right)^4=\frac{1}{81}\)
b) \(\left(-2\frac{1}{4}\right)^3=\frac{-729}{64}\)
c) \(\left(-0,2\right)^2=\frac{1}{25}\)
d) \(\left(-5,3\right)^0=1\)
Cbht
MV
17 tháng 7 2017
\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\\ =\left(2-1\right)\cdot\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}-\dfrac{1}{2^{99}}\\ =1-\dfrac{1}{2^{99}}< 1\)
Vậy \(B< 1\)
17 tháng 7 2017
\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\)
\(\Rightarrow2B=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)\)
\(\Rightarrow2B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{97}}+\dfrac{1}{2^{98}}\)
\(\Rightarrow2B-B=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{97}}+\dfrac{1}{2^{98}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)\)
\(\Rightarrow B=1-\dfrac{1}{2^{99}}\)
\(\rightarrow B< 1\rightarrowđpcm\)
UI
0
\(A=\frac12+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{100}}\)
Ta có:
\(2A=1+\frac12+\frac{1}{2^2}+\cdots+\frac{1}{2^{99}}\)
\(2A-A=\left(1+\frac12+\frac{1}{2^2}+\cdots+\frac{1}{2^{99}}\right)-\left(\frac12+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{100}}\right)\)
\(A=1-\frac{1}{2^{100}}\)
Mà \(\frac{1}{2^{100}}<1\)
\(\rArr1-\frac{1}{2^{100}}<1\) hay \(A<1\)
Vậy \(A<1\)
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