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\(S=1-2+2^2-2^3+...+2^{2012}-2^{2013}\)
\(\Rightarrow2S=2-2^2+2^3-2^4+...+2^{2013}-2^{2014}\)
\(\Rightarrow2S+S=2-2^2+2^3-...-2^{2014}+1-2^2-2^3+...-2^{2013}\)
\(\Rightarrow3S=1-2^{2014}\)\(\Rightarrow3S-2^{2014}=1-2^{2015}\)
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a \(4S=4+4^2+4^3+...+4^{24}\)
\(S=\frac{4S-S}{3}=\frac{4^{24}-1}{3}\)
b/ Xem lại đề bài\(3S=4^{6x}-1=4^{24}-1\Rightarrow6x=24\Rightarrow x=4\)
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Ta có : S = 1 - 2 + 22 - 23 + .... - 22009 + 22010
=> 2S = 2 - 22 + 23 - 24 + .... - 22010 + 22011
Lấy 2S trừ S theo vế ta có :
2S + S = (2 - 22 + 23 - 24 + .... - 22010 + 22011) + (1 - 2 + 22 - 23 + .... - 22009 + 22010)
3S = 22011 + 1
Khi đó 3S - 22011 = 22011 + 1 - 22011 = 1
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Đặt A=1 + 2 + 22+ 23+ 24 +... + 299 + 2100
=>2A=2 + 22+ 23+ 24 +... + 299 + 2100+2101
=>2A-A=(2 + 22+ 23+ 24 +... + 299 + 2100+2101)-(1 + 2 + 22+ 23+ 24 +... + 299 + 2100)
=>A=2101-1
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Ta có:
\(A=1+2+2^2+...+2^{2002}\)
\(2A=2+2^2+2^3+...+2^{2003}\)
\(2A-A=\left(2+2^2+2^3+...+2^{2003}\right)-\left(1+2+2^2+....+2^{2002}\right)\)
\(A=2^{2003}-1\)
Mà: \(2^{2003}=2^{2003}\)
\(\Rightarrow2^{2003}-1< 2^{2003}\)
\(\Rightarrow A< B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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