【解】
（1） 方程A(k)(X^2)+2A(k+1)X+A(k+2)=0,則其
Δ=4[ A(k+1)^2-A(k)*A(k+2)]
=4[ [A(k)+d]^2-A(k)*[A(k)+2d] ]
=4d^2>0;
所以有實(shí)數(shù)解；
（2） 設(shè)A(k)(X^2)+2A(k+1)X+A(k+2)=0的根為X(k),X(k+1);則：
X(k)+X(k+1)=-2A(k+1)/A(k)=-2[(A(k)+d)/A(k)]=-2[1+d/A(k)];
X(k)*X(k+1)=A(k+2)/A(k)=1+2d/A(k);
所以：
1/[X(k+1)+1]-1/[X(k)+1]
=[ X(k)-X(k+1) ]/[X(k+1)*X(k)+( X(k+1)+X(k) )+1]
=[(X(k)+X(k+1))^2-4X(k)*X(k+1)]^(1/2)/[X(k+1)*X(k)+( X(k+1)+X(k) )+1]
=(4d^2/A(k)^2)^(1/2)/(-d/A(k))
=|2d/A(k)|/[-d/A(k)]
= 2
所以{1/(X(k)+1)}成等差數(shù)列