Toán [Toán 6] Bài này giải sao????

P

phamhuy20011801

Đặt $A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+ \dfrac{1}{2^{100}}$
Thì $2A=1+\dfrac{2}{2^2}+\dfrac{2}{2^3}+...+ \dfrac{2}{2^{100}}$
$=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+ \dfrac{1}{2^{99}}$
$2A-A=(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+ \dfrac{1}{2^{99}})-(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+ \dfrac{1}{2^{100}})$
$\rightarrow A=1-\dfrac{1}{2^{100}} < 1$
Vậy...
 
S

sonsuboy

Đặt $S=\dfrac{1}{2}+\dfrac{1}{2^2}..........+ \dfrac{1}{2^{100}}$
--->$2S=1+\dfrac{1}{2}+..........+ \dfrac{1}{99}$
$2S-S= S=1-\dfrac{1}{2^{100}} < 1$
 
B

baobadao2512

$A=\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{100}}$
$ \rightarrow 2A= \dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}+\dfrac{1}{2^{101}} $
$2A-A=1-2^{101}$
$\rightarrow A<1$
 
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T

thanhk12

A=[tex]\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{100}}[/tex]
[tex]\Rightarrow[/tex] 2A=1+ [tex]\frac{1}{2}+....+\frac{1}{2^{99}}[/tex]
[tex]\Rightarrow[/tex] 2A-A=1-[tex]\frac{1}{2^{100}}[/tex]
[tex]\Rightarrow[/tex] A=1-[tex]\frac{1}{2^{100}}[/tex]<1
[tex]\Rightarrow[/tex] A<1
Vậy A<1
 
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C

chi254

Đặt $A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+ \dfrac{1}{2^{100}}$
$A=\dfrac{1}{2} + \dfrac{1}{2}.(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+ \dfrac{1}{2^{99}})$
$A=\dfrac{1}{2} + \dfrac{1}{2}.( A - \dfrac{1}{2^{100}})$
$A=\dfrac{1}{2} +\dfrac{A}{2} - \dfrac{1}{2^{101}}$
$A - \dfrac{A}{2} = \dfrac{1}{2} - \dfrac{1}{2^{101}}$
$\dfrac{A}{2}= \dfrac{1}{2} - \dfrac{1}{2^{101}}$
$\dfrac{A}{2} < \dfrac{1}{2}$
\Rightarrow A < 1
$Vậy ....$
 
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