已知数列为等差数列公差d不为零an不=0 a(k)x2+2a(k+1)x+a(k+2)=0若方程不同的根依次为x1 x2 x3 .xn
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.求证1/（x1+1） 1/(x2+1) 1/(x3+1) .1/(xn+1).是等差数列
证明：∵{an}为等差数列公差d≠0,an≠0
∴2a(n+1)=an+a(n+2)
∴方程anx2+(an+a(n+2))x+a(n+2)=0的二个不同的根分别为：
X1=-an-a(n+2)-an+a(n+2))/(2an)=-1,x2=-2a(n+2)/(2an)= -a(n+2)/an
当方程系数为a1,a2,a3时,x1=-a3/a1
当方程系数为a2,a3,a4时,x2=-a4/a2
当方程系数为a3,a4,a5时,x3=-a5/a3
当方程系数为a4,a5,a6时,x4=-a6/a4
……
当方程系数为an,a(n+1),a(n+2)时,xn=-a(n+2)/an
Bn=1/(xn+1)=1/(1-a(n+2)/an)=an/(an-a(n+2))= -an/(2d)
B(n-1)=a(n-1)/(a(n-1)-a(n+1))= -a(n-1)/(2d)
B(n+1)=a(n+1)/(a(n+1)-a(n+3)) = -a(n+1)/(2d)
B(n+1)-bn= -a(n+1)/(2d)+an/(2d)=-1/2
Bn-b(n-1)= -an/(2d)+ a(n-1)/(2d)=-1/2
即数列{bn}前后项之差为常数
∴数列{bn}为等差数列.