数列｛an｝为等差数列，d≠0，an≠0(n∈正N)，关于x的方程akx2+2ak+1x+ak+2=0

显示全部楼层 · 2010-7-18 15:06:45

①求证当k取不同的正整数时方程有公共根②若方程不同的根依次为x1,x2,x3,...,xn,...求证1/x1+1,1/x2+1,1/x3+1,...,1/xn+1,...是等差数列

千问 · 2010-7-18 15:06:45

是“ak*x^2+2a(k+1)*x+a(k+2)=0”吗？ ①ak*x^2+2a(k+1)*x+a(k+2)=0a(k-1)*x^2+2ak*x+a(k+1)=0两式相减：[ak-a(k-1)]*x^2+2[a(k+1)-ak]*x+[a(k+2)-a(k+1)]=0d*x^2+2d*x+d=0因d≠0x^2+2x+1=0(x+1)^2=0x=-1(公共根)证毕。② ak*x^2+2a(k+1)*x+a(k+2)=0xk+(-1)=-2a(k+1)/akxk*(-1)=a(k+2)/akxk=-a(k+2)/akx1=-a3/a1=-(a1+2d)/a1=-1+2d/a1