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có : Q = [ 2 + 2^2 ] + [ 2^3 +2^4] + ... + [2^9 + 2^10]
Q = 2 [1+2] +2^3[1 +2]+ ...+ 2^9 [1+2]
Q = 2 . 3+2^3 .3 +... + 2^9 .3
Q = 3. [ 2 + 2^3 +... + 2^9]
Vậy Q chia hết cho 3
Ta có:
A = 1 + 2 + 22 + ... + 22008
=> 2A = 2 + 22 + ... + 22009
=> 2A - A = 22009 - 1
=> A = 22008 - 1 < 22009 = B
Vậy B> A
2A=2+2^2+...+2^2009
2A-A=(2+2^2+...+2^2009)-(1+2+...+2^2008)
A=2^2009-1
=>A<B
a) Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}\) ; \(\frac{1}{3^2}< \frac{1}{2.3}\) ; \(\frac{1}{4^2}< \frac{1}{3.4}\) ; ... ; \(\frac{1}{2010^2}< \frac{1}{2009.2010}\)
=> \(Vt< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(=1-\frac{1}{2010}< 1\)
2 . A = 2 + 2^2 +....+2^2009
A = 2A -A = 2^2009-1
A<B
B- A = 1
Ta có:
A = 1 + 2 + 22 + ... + 22008
=> 2A = 2 + 22 + ... + 22009
=> 2A - A = 22009 - 1
=> A = 22009 - 1
Ta có : A = 22009 - 1; B = 22009
=> A - B = 22009 - 1 - 22009 = -1
\(A=1+2+2^2+2^3+.....+2^{2008}\)
\(\Rightarrow2A=2+2^2+2^3+......+2^{2009}\)
\(\Rightarrow2A-A=\left(2+2^2+2^3+......+2^{2009}\right)-\left(1+2+2^2+.....+2^{2008}\right)\)
\(\Rightarrow A=2^{2009}-1\)
\(\Rightarrow A-B=2^{2009}-1-2^{2009}=-1\)
Ta có:
\(A=1+2+2^2+2^3+...+2^{2008}\)
\(\Rightarrow2A=2+2^2+2^3+2^4+...+2^{2009}\)
\(\Rightarrow2A-A=2^{2009}-1\Rightarrow A=2^{2009}-1\)
\(\Rightarrow B-A=2^{2009}-\left(2^{2009}-1\right)=1\)