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a)A=3^0+3^1+3^2+3^3+...+3^2012
A=1+3+3^2+3^3+..+3^2012
3A=3+3^2+3^3+3^4+..+3^2013
3A-A=3+3^2+3^3+3^4+..+3^2013-1-3-3^2-3^3-...-3^2012
2A=3^2013-1
A=\(\frac{3^{2013}-1}{2}\)
B=3^2013
=> A>B
b) A=1+5+5^2+5^3+..+5^99+5^100
5A=5+5^2+5^3+5^4+...+5^100+5^101
5A-A=5+5^2+5^3+5^4+..+5^100+5^101-1-5-5^2-5^3-..-5^99-5^100
4A=5^101-1
A=\(\frac{5^{101}-1}{4}\)
B=5^101/4
=> A<B
a) \(A=2+2^2+2^3+...+2^{2019}\)
\(\Rightarrow2A=2^2+2^3+...+2^{2020}\)
\(\Rightarrow2A-A=\left(2^2+...+2^{2020}\right)-\left(2+...+2^{2019}\right)\)
\(\Rightarrow A=2^{2020}-2\)
Ta có: \(A+2=2^{x+10}\)
\(\Leftrightarrow2^{2020}-2+2=2^{x+10}\)
\(\Leftrightarrow2^{2020}=2^{x+10}\)
\(\Leftrightarrow2020=x+10\)
\(\Leftrightarrow x=2010\)
b) Ta có: \(A+2=2^{2020}=\left(2^{1010}\right)^2\)là số chính phương
XÉT:\(A=2+2^2+2^3+...+2^{2019}\)
\(\Leftrightarrow2A=2^2+2^3+...+2^{2019}+2^{2020}\)
\(\Leftrightarrow2A-A=2^{2020}-2\)
\(\Leftrightarrow A=2^{2020}-2\)
\(\Rightarrow A+2=2^{2020}-2+2=2^{2020}\)LÀ SỐ CHÍNH PHƯƠNG
MÀ\(a+2=2^{x+10}\)
\(\Leftrightarrow2^{x+10}=2^{2020}\)
\(\Leftrightarrow x+10=2020\Leftrightarrow x=2010\)
Ta có A = 20112012 - 20112011 = 20112011 . (2011 - 1) = 20112011 . 2010
B= 20112013 - 20112012 = 20112012 . (2011 - 1) = 20112012 . 2010
Vì 20112012 >20112011 nen 20112011 . 2010 < 20112012 . 2010 hay A<B
A=20112012-20112011
A=20112011.(2011-1)
A=20112011.2010
B=20112013-20112012
B=20112012(2011-1)
B=20112012.2010
=)B>A
2. b)
B = 3 + 32 + 33 + ... + 32016
3B = 32 + 33 + ... + 32017
3B - B = ( 32 + 33 + ... + 32017 ) - ( 3 + 32 + 33 + ... + 32016 )
2B = 32017 - 3
=> 2B + 3 = 32017 - 3 + 3 = 32017 ( đpcm )
3.
a) 3500 = ( 35 )100 = 243100
7300 = ( 73 )100 = 343100
2A=2^2+2^3+2^4+....+2^2014
=>2A-A=(2^2+2^3+2^4+...+2^2014)-(2+2^2+2^3+...+2^20130
=>A=2^2014-2 <2^2014=B
=>A<B
2A=2^2+2^3+...+2^2014
2A+2=2+2^2+...+2^2013+2^2014=A+2^2014
2A-A=2^2014-2
A=2^2014-2<2^2014
Do đó, A<B
Ta có: \(\hept{\begin{cases}3\equiv1\left(mod2\right)\\11\equiv1\left(mod2\right)\end{cases}}\Leftrightarrow\hept{\begin{cases}3^{2013}\equiv1\left(mod2\right)\\11^{2012}\equiv1\left(mod2\right)\end{cases}}\)
\(\Rightarrow A=3^{2013}-11^{2012}\equiv1-1\equiv0\left(mod2\right)\)
=> A chẵn
Mà \(2^{1993}\) và \(4^{71}\) đều chẵn
=> \(2^{1993}+4^{71}\) chẵn