假如有递推数列Xn+1=aXn+bXn-1.在方程两边同时减去yXn,得Xn+1-yXn=(a-y)Xn-Xn-1＝（a-y)(Xn+b/(a-y))我们选择合适的y,令Yn=Xn+1-yXn成为等比数列.这时y只要满足条件－y=b/(a-y)即yy-ay-b=0,解开这个方程,就可以得到...