Question

证明对于任意正整数n,多项式(n+7)²-(n-5)²能被24整除

Answer 1

(n+7)²-(n-5)²=(n+7+n-5)(n+7-n+5)=(2n+2)*12=24(n+1)
所以对于任意正整数n,多项式(n+7)²-(n-5)²能被24整除

Answer 2

(n+7)²-(n-5)²
=(n+7+n-5)(n+7-n+5)
=12(2n+2)
=24(n+1)
结果是24 的倍数，
所以，对于任意正整数n,多项式(n+7)²-(n-5)²能被24整除

Answer 3

(n+7)²-(n-5)²
=n²＋14n＋49－﹙n²－10n＋25﹚
=n²＋14n＋49－n²＋10n－25
=24n＋24
=24﹙n＋1﹚
∴对于任意正整数n,多项式(n+7)²-(n-5)²能被24整除

Answer 4

(n+7)²-(n-5)²=(n+7+n-5)(n+7-n+5)=(2n+2)*12=2*(n+1)*12=24(n+1)
因为n为任意正整数，所以24(n+1)能被24整除。即多项式(n+7)²-(n-5)²能被24整除

Answer 5

(n+7)2-(n-5)2
=(n+7+n-5)(n+7-n+5)
=(2n+2)*12
=24(n+1)....................(**)
(**)/24=(n+1)
故对于任意正整数n,多项式(n+7)²-(n-5)²能被24整除
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