當(dāng)n=2時(shí),3^2>2+3,成立；
設(shè)當(dāng)n=k時(shí),3^k>k+3成立,
當(dāng)n+k+1時(shí),3^(k+1)=3^k * 3>(k+3) * 3=[(k+1)+3]+(2k+5)]>k+1)+3;
綜上所訴,對(duì)于大于1的整數(shù)n 有3的n次方>n+3恒成立.