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题目
探究:2221=2×211×21=21×(21)=212^{2}-2^{1}=2\times 2^{1}-1\times 2^{1}=2^{1}\times \left(2-1\right)=2^{1}
2322=2×221×22=22×(21)=222^{3}-2^{2}=2\times 2^{2}-1\times 2^{2}=2^{2}\times \left(2-1\right)=2^{2}
2423=2×231×23=23×(21)=232^{4}-2^{3}=2\times 2^{3}-1\times 2^{3}=2^{3}\times \left(2-1\right)=2^{3}\ldots \ldots
(1)(1)请你找规律,写出第nn个等式;
(2)(2)计算:27262524232222^{7}-2^{6}-2^{5}-2^{4}-2^{3}-2^{2}-2
(3)(3)计算:1+2+22++22016+22017220181+2+2^{2}+\ldots \ldots +2^{2016}+2^{2017}-2^{2018}.
知识点:解分式方程、分式的加减法、规律型:数字的变化类章节:未标注

答案与解析

答案

(1)2n+12n=2×2n1×2n=2n×(21)=2n\left(1\right)2^{n+1}-2^{n}=2\times 2^{n}-1\times 2^{n}=2^{n}\times \left(2-1\right)=2^{n}
(2)(2)由(1)的结论得:
27262524232222^{7}-2^{6}-2^{5}-2^{4}-2^{3}-2^{2}-2
=26252423222=2^{6}-2^{5}-2^{4}-2^{3}-2^{2}-2
=252423222=2^{5}-2^{4}-2^{3}-2^{2}-2
=2423222=2^{4}-2^{3}-2^{2}-2
=23222=2^{3}-2^{2}-2
=222=2^{2}-2
=2=2
(3)(3)原式=(220182201722016220152221)=-(2^{2018}-2^{2017}-2^{2016}-2^{2015}-\ldots \ldots -2^{2}-2-1)
=1=-1

解析

(1)2n+12n=2×2n1×2n=2n×(21)=2n\left(1\right)2^{n+1}-2^{n}=2\times 2^{n}-1\times 2^{n}=2^{n}\times \left(2-1\right)=2^{n}
(2)(2)由(1)的结论得:
27262524232222^{7}-2^{6}-2^{5}-2^{4}-2^{3}-2^{2}-2
=26252423222=2^{6}-2^{5}-2^{4}-2^{3}-2^{2}-2
=252423222=2^{5}-2^{4}-2^{3}-2^{2}-2
=2423222=2^{4}-2^{3}-2^{2}-2
=23222=2^{3}-2^{2}-2
=222=2^{2}-2
=2=2
(3)(3)原式=(220182201722016220152221)=-(2^{2018}-2^{2017}-2^{2016}-2^{2015}-\ldots \ldots -2^{2}-2-1)
=1=-1

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