氟代丙烷 猛将:求证:11…1(2n个)-22…2(n个)是一个完全平方数
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初二数学
A = 11...1(2n个)
= 1 + 10 + 100 + ..... + 10^(2n-1)
= [10^(2n) - 1]/9
B = 22...2(n个)
= 2*1 + 2*100 + .... + 2*10^(n-1)
= 2*[10^n - 1]/9
A - B
= [10^(2n) - 2*10^n + 1]/9
= [(10^n)^2 - 2*10^n + 1]/9
= [(10^n - 1)^2]/9
= [(10^n - 1)/3]^2
= [33....3(n个)]^2
11…1(2n个)-22…2(n个)
=11…1(n个)00...0(n个)+11...1(n个)-22…2(n个)
=11...1(n个)*(10^n)-11...1(n个)
=11...1(n个)*(10^n-1)
=11...1(n个)*99...9(n个)
=11...1(n个)*11...1(n个)*3*3
=(33...3)^2
注a^b表示a的b次方
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