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求证:N=(5^2)*(3^2n+1)*(2^n)-(3^n)*(6^n+2)
人气:316 ℃ 时间:2020-05-12 23:05:52
解答
是不是求证这个多项式能被13整除?
N=(5^2)*(3^2n+1)*(2^n)-(3^n)*(6^n+2)
=5^2*3^2n+1*2^n-3^n*(2*3)^n+2
=5^2*3^2n+1*2^n-3^n*2^n+2*3^n+2
=5^2*3^2n+1*2^n-3^n*2^n*2^2*3^n+1*3
=5^2*(3^2n+1*2^n)-(3^2n+1*2n)*2^2*3
=(25-12)*(3^2n+1*2^n)
=13*(3^2n+1*2^n)
我刚才就在做这道题……
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