1.当N=5时,2^N=32,N^2=25,2^N>N^22.假定当N=k时不等式成立,即：2^k-k^2>0,则当N=k+1时2^N-N^2=2*2^k-(k+1)^2=(2^k-k^2)+(2^k-2k-1)>k^2-2k-1=(k-1)^2-2,因为：k>5,所以(k-1)^2-2>0于是,2*2^k-(k+1)^2>0即：若N=k时,2...