解:(1)由題可知:
原式=(2-1)(2
5+2
4+2
3+2
2+2+1)=2
6-1=64-1=63;
(2)原式=(2-1)(2
2011+2
2010+2
2009+2
2008+…+2+1…)=2
2012-1,
∵2
1=2,2
2=4,2
3=8,2
4=16,2
5=32,2
6=64,
∴2
n(n為自然數(shù))的各位數(shù)字只能為2,4,8,6,且具有周期性.
∴2012÷4=503×4,
∴2
2011+2
2010+2
2009+2
2008+…+2+1的個(gè)位數(shù)字是6-1=5;
(3)設(shè)S=
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+
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+
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+…+
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+
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,
則2S=1+
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+
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+
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+…+
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,
所以,S=1-
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.
分析:(1)根據(jù)已知(x-1)(x
3+x
2+x+1)=x
4-1,得出原式=(2-1)(2
5+2
4+2
3+2
2+2+1)求出即可;
(2)根據(jù)已知(1)中所求,求出2
n(n為自然數(shù))的各位數(shù)字只能為2,4,8,6,且具有周期性,進(jìn)而求出答案;
(3)根據(jù)已知得出1-
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=
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,
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=
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-
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,
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=
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-
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,進(jìn)而求出即可.
點(diǎn)評(píng):此題主要考查了數(shù)字的變化規(guī)律;根據(jù)已知得出數(shù)字變化與不變是解決本題的突破點(diǎn).