（3n+1)(3n-1)-(3-n)(3+n)
=9n^2-1-(9-n^2)
=9n^2-1-9+n^2
=10n^2-10
=10(n^2-1)是10的倍數(shù).
n=1時(shí),（3n+1)(3n-1)-(3-n)(3+n)=0,是10的倍數(shù).0是10的倍數(shù)？？0當(dāng)然是10的倍數(shù)。10的0倍
對(duì)于任意的正整數(shù)n,試說(shuō)明整數(shù)(3n+1)(3n-1)-(3-n)(3+n)的值一定是10的倍數(shù)
對(duì)于任意的正整數(shù)n,試說(shuō)明整數(shù)(3n+1)(3n-1)-(3-n)(3+n)的值一定是10的倍數(shù)
請(qǐng)說(shuō)明當(dāng)N=1時(shí),
請(qǐng)說(shuō)明當(dāng)N=1時(shí),
數(shù)學(xué)人氣：219 ℃時(shí)間：2019-08-20 01:35:46
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