设有数列an,a1=5/6,若以a1,a2,...an为系数的二次方程:an-1X^2-anx+1=0,都有根A,B,且3A-AB+3B=1求证{an-1/2}为等比数列
问题描述：
设有数列an,a1=5/6,若以a1,a2,...an为系数的二次方程:an-1X^2-anx+1=0,都有根A,B,且3A-AB+3B=1
求证{an-1/2}为等比数列
答
A+B = An/An-1
AB = 1/An-1
3(An/An-1) - 1/An-1 =1
=>
3An - 1 =An-1
3(An - 1/2) = An-1 - 1/2
故q = 1/3
答
因为an-1X^2-anx+1=0,所以A+B=an/an-1 AB=1/an-1 又因为3A-AB+3B=1,
所以n大于等于2时
（3an -1)/an-1=1 即3an-1=an-1所以
（an-1/2）/（an-1 -1/2）=1/3
所以{an-1/2}为等比数列