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人教初中代数第二册P4 7有这样一道题 :求证 :当n是整数时 ,两个连续奇数的平方差 ( 2n + 1 ) 2 -( 2n - 1 ) 2 是 8的倍数。分析 :利用平方差公式进行因式分解 ,可以证明。证明 :( 2n + 1 ) 2 - ( 2n - 1 ) 2 =[( 2n + 1 ) + ( 2n - 1 ) ][( 2n + 1 ) - ( 2n - 1 ) ]
People teach junior high school algebra second volume P4 7 to have such a question: Proof: When n is an integer, the square difference (2n + 1) 2 - (2n - 1) 2 of 2 consecutive odd numbers is a multiple of 8. Analysis: Using the square difference formula for factorization can be proved. Proof: (2n + 1) 2 - (2n - 1) 2 = [(2n + 1) + (2n - 1)] [(2n + 1) - (2n - 1)]