Question

已知等差数列{an}的公差d不等于0,对于任给的n属于正整数都有an不等于0,求证：（1）对任给的n属于正整数,所有方程

Answer 1

an x^2 + 2 (an+d) x+ an+2d =0

an(x^2+2x+1)+2d(x+1)=0

an(x+1)^2+2d(x+1)=0

an(x+1)(x+1+2d)=0

所以有个根为x1=-1-2d 与n无关.

x1+x2=-b/a

x1x2=c/a

-1-2d+an=-(2an+2d) / (an)

an^2+(1-2d)an+2d=0

(-1-2d)an=(an+2d)/an

(1+2d)an^2+an+2d=0

然后an没求出来.自己再看看吧