证明当k≥4时2^（k-1）＞k+2

用导数证明的设f(k)=2^(k-1),g(k)=k+2f'(k)=2^(k-1)*ln2g'(k)=1k≥4时f'(k)=2^(k-1)*ln2是增函数∴f'(k)最小值=2^3*ln2=8ln2>1∴k≥4时f'(x)>g'(k)∵f(4)=2^3=8>g(4)=6∴f(k)>g(k)即2^（k-1）＞k+2