∵{an}是等差數(shù)列,∴2a(r+1)=a(r)+a(r+2),即a(r)- 2a(r+1)+a(r+2)=0
故當(dāng)x=-1時(shí),a(r)x^2+2a(r+1)x+a(r+2)
= a(r)- 2a(r+1)+a(r+2)=0
∴當(dāng)r取不同自然數(shù)時(shí),原方程有一個(gè)公共根－1還有一問：設(shè)另一根為m(r),求證：1/(m(1)+1),1/(m(2)+1),......,1/(m(n)+1).........也是等差數(shù)列在說下公共根是什么意思。 會(huì)加分的。a(r)X²+2a(r+1)X+a(r+2)=0(r屬于N+),公共根就是指當(dāng)r=1時(shí)，方程有這個(gè)根；當(dāng)r=2時(shí)，方程也有這個(gè)根；當(dāng)r=3時(shí)，方程也有這個(gè)根；………………ar*x^2+2a(r+1)*x+a(r+2)=0由第①小題可知：方程有一個(gè)根-1，設(shè)另一個(gè)根是mr，根據(jù)韋達(dá)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).........是等差數(shù)列，公差為-1/2.good