等差数列｛an｝中已知公差d不等于0,an不等于0,设方程

∵{an}是等差数列,∴2a(r+1)=a(r)+a(r+2),即a(r)- 2a(r+1)+a(r+2)=0
故当x=-1时,a(r)x^2+2a(r+1)x+a(r+2)
= a(r)- 2a(r+1)+a(r+2)=0
∴当r取不同自然数时,原方程有一个公共根－1还有一问：设另一根为m(r),求证：1/(m(1)+1),1/(m(2)+1),......,1/(m(n)+1).........也是等差数列在说下公共根是什么意思。 会加分的。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)/armr*(-1)=a(r+2)/armr=-a(r+2)/arm1=-a3/a1=-(a1+2d)/a1=-1+2d/a1m(r-1)=-a(r+1)/a(r-1)1+mr=1-a(r+2)/ar=-2d/ar1/[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.good