So sánh 1+22+23+...+22005 với 22006
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mik bt lm câu 1 thôi nha, bn thông cảm:
a = 2007.2009 b = 20082
=(2008 - 1)(2008 + 1)
= 20082 - 1
Ta có, a = 20082 - 1, b = 20082
mà 20082 - 1 < 20082
=> a < b
\(S=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2005}}\)
\(2.S=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\)
\(2.S-S=\left(2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\right)\)
\(S=2-\dfrac{1}{2^{2006}}\)
\(A=4+2^2+2^3+...+2^{2006}\)
\(\mathsf{Đặt}:B=2^2+2^3+...+2^{2006}\\2B=2^3+2^4+...+2^{2007}\\2B-B=(2^3+2^4+...+2^{2007})-(2^2+2^3+...+2^{2006})\\B=2^{2007}-2^2\\B=2^{2007}-4\)
Thay \(B=2^{2007}-4\) vào A, ta được:
\(A=4+(2^{2007}-4)\\\Rightarrow A=2^{2007}\)
$\Rightarrow A$ là 1 luỹ thừa của cơ số 2.
Vậy: ...
Đặt A=22+23+..+22005
2A=23+24+..+22006
suy ra 2A-A=(23+24+..+22006) - (22+23+..+22005)
A=22006-22
suy ra C=4+22006-4
C=22006 .Là lũy thừa của 2 (đpcm)
2A=2*(1+2+22+...+22020)=2+22+...+22021
2A-A=(1+2+22+...+22021)-(1+2+22+...+22020)
A=22021-1<2021
Giải:
A=1+2+22+23+...+22020
2A=2+22+23+24+...+22021
2A-A=(2+22+23+24+...+22021)-(1+2+22+23+...+22020)
A=22021-1
⇒A<22021
Chúc bạn học tốt!
S=1+2+22+...+29�=1+2+22+...+29
2S=2(1+2+22+...+210)2�=2(1+2+22+...+210)
2S=2+22+23+...+292�=2+22+23+...+29
2S−S=(2+22+23+...+210)−(1+2+22+...+29)2�−�=(2+22+23+...+210)−(1+2+22+...+29)
\(S=2^{10}-1=2^2.2^8-1=4.2^8-1
HT
S=1+2+22+...+29�=1+2+22+...+29
2S=2(1+2+22+...+210)2�=2(1+2+22+...+210)
2S=2+22+23+...+292�=2+22+23+...+29
2S−S=(2+22+23+...+210)−(1+2+22+...+29)2�−�=(2+22+23+...+210)−(1+2+22+...+29)
\(S=2^{10}-1=2^2.2^8-1=4.2^8-1
Có : \(S=1+2+2^2+2^3+....+2^{99}\)
\(\Rightarrow2S=2+2^2+2^3+....+2^{100}\)
\(\Rightarrow2S-S=\left(2+2^2+2^3+...+2^{100}\right)-\left(1+2+2^2+....+2^{99}\right)\)
\(\Rightarrow S=2^{100}-1< 2^{100}\)
Vậy \(S< 2^{100}\)
S=1+2+22+23+....+299
⇒2S=2+22+23+....+2100
⇒2S−S=2100-1
S=2100-1
vì 2100 -1<2100
⇒S<2100
\(S=1+2+2^2+2^3+...+2^9\)
Đặt \(2S=2+2^2+2^3+2^4+...+2^{10}\)
\(2S-S=2^{10}-1\) hay \(S=2^{10}-1< 2^{10}\)
\(\Rightarrow\) \(2^{10}=2^2.2^8< 5.2^8\)
Vậy \(S< 5.2^8\)
\(#Tuyết\)
2S=2+2^2+...+2^10
=>S=2^10-1=1023
5*2^8=256*5=1280
=>S<5*2^8
Sửa đề: \(S=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{50}\)
Ta có: \(S=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{50}\)
\(=\dfrac{1}{20}+\left(\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{30}\right)+\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)\)
\(\Leftrightarrow S>\dfrac{1}{20}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{4}\)
\(\Leftrightarrow S>\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{3}{4}\)(đpcm)
1 + 22 + 23 + ... + 22005
Gọi dãy số trên là A
A = \(1+2^2+2^3+....+2^{2005}\)
A =\(2^0+2^2+2^3+....+2^{2005}\)
A + \(2^1\)= \(2^0+2^1+2^2+2^3+....+2^{2005}\)
( A + 2 ) x 21 = \(\left(2^0+2^1+2^2+2^3+....+2^{2005}\right)\times2^1\)
Ax2 + 4 =\(2^1+2^2+2^3+2^4+....+2^{2006}\)
4 + A x 2 - A =\(2^1+2^2+2^3+2^4+....+2^{2006}-\left(1+2^2+2^3+...2^{2005}\right)\)
4 + A = \(2^1+2^2+2^3+2^4+....+2^{2006}-1-2^2-2^3-....-2^{2005}\)
4 + A = \(2^{2006}-1\)
A=\(2^{2006}-1-4\)
A = \(2^{2006}-5\)
Mà \(2^{2006}-5< 2^{2006}\)
\(\Rightarrow1+2^2+2^3+....+2^{2005}< 2^{2006}\)