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a ) \(A=2^0+2^1+2^2+...+2^{2010}\)
\(\Rightarrow2A=2+2^2+2^3+...+2^{2011}\)
\(\Rightarrow2A-A=\left(2+...+2^{2011}\right)-\left(2^0+2^1+...+2^{2010}\right)\)
\(\Rightarrow2A-A=2^{2011}-2^0\)
\(\Rightarrow A=2^{2011}-1\)
b ) \(B=1+3+3^2+...+3^{100}\)
\(\Rightarrow3B=3+3^2+3^3+...+3^{101}\)
\(\Rightarrow3B-B=\left(3+3^2...+3^{2011}\right)-\left(1+3+...+3^{2010}\right)\)
\(\Rightarrow2B=3^{2011}-1\)
\(\Rightarrow B=\frac{3^{2011}-1}{2}\)
Chúc bạn học tốt !!!
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c)ta thấy 125*5^12=(5^3)*(5^12)=5^15 , =>124*5^12 <5^15
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\(A=1+2^2+2^4+2^6+...+2^{100}\)
\(4A=2\left(1+2^2+2^4+2^6+...+2^{100}\right)=2+2^4+2^6+2^8+...+2^{100}+2^{102}\)
\(4A-A=\left(2^2+2^4+2^6+2^8+...+2^{100}+2^{102}\right)-\left(1+2^2+2^4+...+2^{100}\right)\)
\(3A=2^{102}-1\)
\(A=\frac{2^{102}-1}{3}\)
\(B=2+2^3+2^5+2^7+...+2^{1001}\)
\(4B=2^3+2^5+2^7+...+2^{1001}+2^{1003}\)
\(4B-B=\left(2^3+2^5+2^7+...+2^{1001}+2^{1003}\right)-\left(2+2^3+2^5+...+2^{1001}\right)\)
\(3B=2^{1003}-2\)
\(B=\frac{2^{1003}-2}{3}\)
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1.A=2^2+2^4+...+2^2010
=> 2^2 A= 2^4+2^6+..+2^2012
=> 2^2 A - A=( 2^4+2^6+..+2^2012 ) -(2^2+2^4+...+2^2010 )
=> 3A= 2^2012 -2^2
=> A= (2^2012-2^2)/3
B=3-3^2+3^3-...-3^2010
=>3B= 3^2 -3^3+3^4-...-3^2011
=> 3B + B = (3^2 -3^3+3^4-...-3^2011) +(3-3^2+3^3-...-3^2010)
=> 4B =3-3^2011
=> B= (3-3^2011)/4
2.
A=3+3^2+..+3^100
=> 3A =3^2+3^3+...+3^101
=> 3A- A = (3^2+3^3+...+3^101)-(3+3^2+..+3^100)
=> 2A=3^101 -3
=> 2A+3 =3^101 mà 2A+3 =3^n
=> n=101
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a)Ta có \(2A=2^2+2^3+...+2^{101}\)
\(\Rightarrow2A-A=\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+2^3+...+2^{100}\right)\)
\(\Rightarrow A=2^{101}-2\)
Vậy \(A=2^{101}-2\)
b)
Ta có \(3A=3^2+3^3+...+3^{101}\)
\(\Rightarrow3A-A=\left(3^2+3^3+...+3^{101}\right)-\left(3+3^2+3^3+...+3^{100}\right)\)
\(\Rightarrow2A=3^{101}-3\)
\(\Rightarrow A=\frac{3^{101}-3}{2}\)
Vậy \(A=\frac{3^{101}-3}{2}\)
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