cho A = 1+2+22+...........22009 và B= 22010 -1
So sánh A và B
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Ta có: A = 1 + 2 + 2 2 + . . . + 2 2009 + 2 2010
= 1 + 2 ( 1 + 2 + 2 2 ) + ... + 2 2008 ( 1 + 2 + 2 2 )
= 1 + 2 ( 1 + 2 + 4 ) + ... + 22008 ( 1 + 2 + 4 )
= 1 + 2 . 7 + ... + 2 2008 . 7 = 1 + 7 ( 2 + ... + 2 2008 )
Mà 7 ( 2 + ... + 2 2008 ) ⋮ 7. Do đó: A chia cho 7 dư 1.
Ta có: A = 1 + 2 + 2 2 + 2 3 + ... + 2 2008 + 2 2009 + 2 2010
= 1 + 2 ( 1 + 2 + 22 ) + ... + 2 2008 ( 1 + 2 + 22 )
= 1 + 2 ( 1 + 2 + 4 ) + ... + 2 2008 ( 1 + 2 + 4 )
= 1 + 2 . 7 + ... + 2 2008 . 7 = 1 + 7 ( 2 + ... + 2 2008 )
Mà 7 ( 2 + ... + 2 2008 ) ⋮ 7. Do đó: A chia cho 7 dư 1.
A = 2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰¹⁰
⇒ 2A = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰¹¹
⇒ A = 2A - A = (2 + 2² + 2³ + 2⁴ + ... + 2²⁰¹¹) - (2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰¹⁰)
= 2²⁰¹¹ - 2⁰
= 2²⁰¹¹ - 1
= B
Vậy A = B
\(A=1+2+2^2+2^3+...+2^{2021}\)
\(\Rightarrow2A=2+2^2+2^3+...+2^{2022}\)
\(\Rightarrow A=2A-A=2+2^2+...+2^{2022}-1-2-2^2-...-2^{2021}=2^{2022}-1>2^{2021}-1=N\)
\(a=1+2+2^2+...+2^{2021}\\ \Rightarrow2a=2+2^2+2^3+...+2^{2022}\\ \Rightarrow2a-a=\left(2+2^2+2^3+...+2^{2022}\right)-\left(1+2+2^2+...+2^{2021}\right)\\ \Rightarrow a=2^{2022}-1>2^{2021}-1=n\)
Bài 1:
\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)
\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)
Bài 2:
\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)
\(10A=10.\dfrac{10^{2004}+1}{10^{2005}+1}=\dfrac{10^{2005}+10}{10^{2005}+1}=1+\dfrac{9}{10^{2005}+1}\\ 10B=10.\dfrac{10^{2005}+1}{10^{2006}+1}=\dfrac{10^{2006}+10}{10^{2006}+1}=1+\dfrac{9}{10^{2006}+1}\)
vì \(\dfrac{9}{10^{2005}+1}>\dfrac{9}{10^{2006}+1}\Rightarrow10A>10B\Rightarrow A>B\)
Đặt A = 22009 + 22008 + ... + 21 + 20. Khi đó, M = 22010 - A
Ta có 2A = 22010 + 22009 + ... + 22 + 21.
Suy ra 2A - A = 22010 - 20 = 22010 - 1.
Do đó M = 22010 - A = 22010 - (22010 - 1) = 22010 - 22010 + 1 = = 1.
M=2^2010-(2^2009+2^2008+2^2007+...+2^1+2^0)
M=22010-22009-22008-22007-...-21-20
=>2M=22011-22010-22009-22008-...-22-21
=>2M-M=22011-22010-22009-22008-...-22-21-(22010-22009-22008-22007-...-21-20)
=>M=22011-22010-22009-22008-...-22-21-22010+22009+22008+22007+...+21+20
=22011-22010-22010+20
=22011-2.22010+1
=22011-22011+1
=1
vậy M=1
\(a,\Rightarrow2A=2+2^2+...+2^{2011}\)
\(\Rightarrow2A-A=2+2^2+...+2^{2011}-2^0-2-..-2^{2010}\)
\(\Rightarrow A=2^{2011}-1=B\)
\(b,A=2019.2011=\left(2010-1\right)\left(2010+1\right)=\left(2010-1\right).2010+\left(2010-1\right)=2010^2-2010+2010-1=2010^2-1< 2010^2=B\)
\(a,\Rightarrow2A=2^1+2^2+...+2^{2011}\\ \Rightarrow2A-A=A=2^{2011}-2^0=2^{2011}-1=B\)
\(b,A=\left(2010-1\right)\left(2010+1\right)=2010^2+2010-2010-1=2010^2-1< 2010^2=B\)
cho A = 1+2+22+...........22009 và B= 22010 -1
2A = 2 + 2^2 + ......+ 2^2010
2A - A = 2^2010 - 1 =A
Vì vậy A = B