(1)
ak*x^2+2a(k+1)*x+a(k+2)=0
ak*x^2+2(ak+d)*x+ak+2d=0
x=-1或-1-2d/ak
公共根为-1
(2)
-1-2d/ak=ak
an+1=-2d/an
1/(an+1)=-an/2d
由an为等差数列,d为常数
所以1/(an+1)为等差数列
得证
(1)
ak*x^2+2a(k+1)*x+a(k+2)=0
ak*x^2+2(ak+d)*x+ak+2d=0
x=-1或-1-2d/ak
公共根为-1
(2)
-1-2d/ak=ak
an+1=-2d/an
1/(an+1)=-an/2d
由an为等差数列,d为常数
所以1/(an+1)为等差数列
得证