举报
玻璃心5202003
a(r)X²+2a(r+1)X+a(r+2)=0(r属于N+), 公共根就是指当r=1时,方程有这个根; 当r=2时,方程也有这个根; 当r=3时,方程也有这个根; ……………… ar*x^2+2a(r+1)*x+a(r+2)=0 由第①小题可知:方程有一个根-1, 设另一个根是mr, 根据韦达定理可知:mr+(-1)=-2a(r+1)/ar mr*(-1)=a(r+2)/ar mr=-a(r+2)/ar m1=-a3/a1=-(a1+2d)/a1=-1+2d/a1 m(r-1)=-a(r+1)/a(r-1) 1+mr=1-a(r+2)/ar=-2d/ar 1/[1+mr]=-ar/(2d) 所以 1/[1+m(r-1)]=-a(r-1)/(2d) 1/[1+mr]-1/[1+m(r-1)]=-ar/(2d)+a(r-1)/(2d) =-[ar-a(r-1)]/(2d) =-d/(2d) =-1/2 所以1/(m(1)+1),1/(m(2)+1),......,1/(m(n)+1).........是等差数列,公差为-1/2.