Cho B = 4 + 32 + 33 +...+32003 +32004 và C = 32005
So sánh B với C
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A = 3 + 32 + 33 + ... + 32004
A = ( 3 + 32 + 33 + 34 ) + ... + ( 32001 + 32002 + 32003 + 32004 )
A = 3( 1 + 3 + 32 + 33 ) + ... + 32001( 1 + 3 + 32 + 39 )
A = 3.40 + ... + 32001.40
A = ( 3 + 35 + ... 32001) . 40
=> A chia hết cho 40
Bài 1:
a. $2^{29}< 5^{29}< 5^{39}$
$\Rightarrow A< B$
b.
$B=(3^1+3^2)+(3^3+3^4)+(3^5+3^6)+...+(3^{2009}+3^{2010})$
$=3(1+3)+3^3(1+3)+3^5(1+3)+...+3^{2009}(1+3)$
$=(1+3)(3+3^3+3^5+...+3^{2009})$
$=4(3+3^3+3^5+...+3^{2009})\vdots 4$
Mặt khác:
$B=(3+3^2+3^3)+(3^4+3^5+3^6)+....+(3^{2008}+3^{2009}+3^{2010})$
$=3(1+3+3^2)+3^4(1+3+3^2)+...+3^{2008}(1+3+3^2)$
$=(1+3+3^2)(3+3^4+....+3^{2008})=13(3+3^4+...+3^{2008})\vdots 13$
Bài 1:
c.
$A=1-3+3^2-3^3+3^4-...+3^{98}-3^{99}+3^{100}$
$3A=3-3^2+3^3-3^4+3^5-...+3^{99}-3^{100}+3^{101}$
$\Rightarrow A+3A=3^{101}+1$
$\Rightarrow 4A=3^{101}+1$
$\Rightarrow A=\frac{3^{101}+1}{4}$
a: \(A=2019\cdot2021=2020^2-1\)
\(B=2020^2\)
Do đó: A<B
Bài 1:
\(2^{49}=\left(2^7\right)^7=128^7;5^{21}=\left(5^3\right)^7=125^7\\ Vì:128^7>125^7\Rightarrow2^{49}>5^{21}\)
Bài 2:
\(a,S=1+3+3^2+3^3+...+3^{99}\\ =\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^3\right)+...+3^{96}.\left(1+3+3^2+3^3\right)\\ =40+3^4.40+...+3^{96}.40\\ =40.\left(1+3^4+...+3^{96}\right)⋮40\\ b,S=1+4+4^2+4^3+...+4^{62}\\ =\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+...+4^{60}.\left(1+4+4^2\right)\\ =21+4^3.21+...+4^{60}.21\\ =21.\left(1+4^3+...+4^{60}\right)⋮21\)
Bài 1 :
\(2^{49}=\left(2^7\right)^7=128^7\)
\(5^{21}=\left(5^3\right)^7=125^7\)
mà \(125^7< 128^7\)
\(\Rightarrow2^{49}>5^{21}\)
Bài 2 :
a) \(S=1+3+3^2+3^3+...3^{99}\)
\(\Rightarrow S=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)...+3^{96}\left(1+3+3^2+3^3\right)\)
\(\Rightarrow S=40+40.3^4+...+40.3^{96}\)
\(\Rightarrow S=40\left(1+3^4+...+3^{96}\right)⋮40\)
\(\Rightarrow dpcm\)
b) \(S=1+4+4^2+4^3+...4^{62}\)
\(\Rightarrow S=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...4^{60}\left(1+4+4^2\right)\)
\(\Rightarrow S=21+4^3.21+...4^{60}.21\)
\(\Rightarrow S=21\left(1+4^3+...4^{60}\right)⋮21\)
\(\Rightarrow dpcm\)
1. So sánh bằng cách nhanh nhất:
a. \(\frac{28}{31}>\frac{28}{33}\)
b.\(\frac{28}{81}>\frac{11}{34}\)
c. \(\frac{13}{11}>\frac{31}{32}\)
d. \(\frac{1994}{1995}>\frac{36}{37}\)
Giúp mình với!!!
A=11+22+33+44
=33+77
=110
B=14+23+32+41
=37+73
=110
VÌ 110=110 NÊN A =B
A = 11 + 22 + 33 + 44
= 33 + 33 + 44
= 66 + 44
= 110
B = 14 + 23 + 32 + 41
= 37 + 73
= 110
Vậy A = B ( = 110 )
a) 18 2 < 10 3
b) 3 2 + 4 2 < ( 3 + 4 ) 2
c) 100 2 + 30 2 < ( 100 + 30 ) 2
d) a 2 + b 2 > ( a - b ) 2 với a ∈ N * ; b ∈ N * .
Ta có :
\(B=4+3^2+3^3+...+3^{2003}+3^{2004}\)
\(B=1+3+3^2+3^3+...+3^{2003}+3^{2004}\)
\(3B=3+3^2+3^3+...+3^{2004}+3^{2005}\)
\(3B-B=\left(3+3^2+3^3+...+3^{2004}+3^{2005}\right)-\left(1+3+3^2+...+3^{2003}+3^{2004}\right)\)
\(2B=3^{2005}-1\)
Vì : \(2B=3^{2005}-1< 3^{2005}=A\)
Nên \(B< A\) hay \(A>B\)
Vậy \(A>B\)
Chúc bạn học tốt ~