n 是一个正整数。说明对于所有的 n， $(7n+2)^{2}-(7n-1)^{2}$ 是否都是 3 的倍数。答案 ___

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完整解法

1. Recognise and apply the difference of two squares identity.

Let $a=7n+2$ and $b=7n-1$ .

$(7n+2)^2-(7n-1)^2=[(7n+2)+(7n-1)][(7n+2)-(7n-1)]$

2. Simplify the two factors.

$(7n+2)+(7n-1)=14n+1$
$(7n+2)-(7n-1)=7n+2-7n+1=3$

So
$(7n+2)^2-(7n-1)^2=(14n+1)(3).$

3. Conclude whether the expression is always a multiple of 3.

$(14n+1)(3)=3(14n+1).$

Since $n$ is a positive integer, $14n+1$ is an integer. Hence $3(14n+1)$ is a multiple of 3.

答案

Yes. Since $(7n+2)^2-(7n-1)^2=3(14n+1)$ , it is a multiple of 3 for all positive integers $n$ .

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