用特值代入验证即可.
【解析】
令a=1,b=2,k=2得到abk+akb=6,ak+1+bk+1=9,故(abk+akb)-(ak+1+bk+1)<0;
令a=2,b=1,k=2得到abk+akb=6,ak+1+bk+1=9,故(abk+akb)-(ak+1+bk+1)<0;
令a=2,b=1,k=1得到abk+akb=4,ak+1+bk+1=5,故(abk+akb)-(ak+1+bk+1)<0;
故(abk+akb)-(ak+1+bk+1) (k∈N*)的符号与与k的奇偶性无关
故答案为 B