Question

设a_n=2ⁿ,b_n=n²,试比较a_n与b_n的大小.

Answer 1

解：当n=1时,a₁＞b₁;当n=2时,a₂=b₂;

当n=3时,a₃＜b₃;当n=4时,a₄=b₄;

当n=5时,a₅＞b₅;当n=6时,a₆＞b₆;

猜想n≥5时,a_n＞b_n,下面用数学归纳法证明,当n≥5时,a_n＞b_n.

(1)当n=5时,由上验证知a_n＞b_n成立.

(2)假设当n=k时,a_k＞b_k成立,

即2^k＞k²,当n=k+1时a_k+1=2^k+1=2·2^k＞2·k²＞k²+2k+1=(k+1)²=b_k+1.

(当n＞5时,k²＞2k+1成立)

∴当n=k+1时,a_k+1＞b_k+1成立.

综上,可知当n≥5时,a_n＞b_n.