Question

當(dāng)n∈N^*時(shí),試比較a_n=2ⁿ與b_n=n²的大小.

      

Answer 1

解析:a₁=2,b₁=1,∴a₁>b₁;?

       a₂=4,b₂=4,∴a₂=b₂;?

       a₃=8,b₃=9,∴a₃<b₃;?

       a₄=16,b₄=16,∴a₄=b₄;?

       a₅=32,b₅=25,∴a₅>b₅;?

       a₆=64,b₆=36,∴a₆>b₆.?

       猜想:n≥5時(shí),a_n>b_n(以下用數(shù)學(xué)歸納法證明當(dāng)n≥5時(shí),a_n>b_n).?

       (1)當(dāng)n=5時(shí),a₅=32,b₅=25,a₅>b₅成立.?

       (2)假設(shè)當(dāng)n=k(k∈N^*,k≥5)時(shí),有a_k>b_k,即2^k>k²(k≥5),?

       因此a_k+1-b_k+1=2^k+1-(k+1)²=2·2^k-(k²+2k+1)>2k²-(k²+2k+1)?

       =k²-2k-1=(k-1)²-2.?

       ∵k≥5,∴(k-1)²≥16.?

       ∴(k-1)²-2>0,即a_k+1>b_k+1?.?

       這表明當(dāng)n=k+1時(shí),a_k+1>b_k+1.?

       根據(jù)(1)(2)可知,當(dāng)n≥5(n∈N^*)時(shí)a_n>b_n.?

       綜上可知,當(dāng)n=2,4時(shí),a_n=b_n;當(dāng)n=3時(shí),a_n<b_n;當(dāng)n=1或n≥5時(shí),a_n>b_n.