已知等差数列{an}中,an≠0,公差d≠0,(1)求证,方程anx^2+2a(n+1)x+2a(n+2),(2)设(1)中方程的另一根为{bn},求证{1/(bn+1)}为等差数列
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已知等差数列{an}中,an≠0,公差d≠0,(1)求证,方程anx^2+2a(n+1)x+2a(n+2),(2)设(1)中方程的另一根为{bn},求证{1/(bn+1)}为等差数列
已知等差数列{an}中,an≠0,公差d≠0,(1)求证,方程anx^2+2a(n+1)x+2a(n+2),
(2)设(1)中方程的另一根为{bn},求证{1/(bn+1)}为等差数列
已知等差数列{an}中,an≠0,公差d≠0,(1)求证,方程anx^2+2a(n+1)x+2a(n+2),(2)设(1)中方程的另一根为{bn},求证{1/(bn+1)}为等差数列
证明
(1)
∵{an}是等差数列,
∴2a(k+1)=ak+a(k+2),
故方程akx^2+2ak+1x+ak+2=0可变为
[akx+a(k+2)](x+1)=0,
∴当k取不同自然数时,原方程有一个公共根-1
(2)原方程另一根为bn=xk=- a(k+2)/ak=(ak+2d)/ak=-1-2d/ak
∴1/(bn+1=1/(xk+1)=-ak/2d
∵1/[x(k+1)+1]-1/(xk+1)=-ak+1/2d-(ak/2d)=(ak-ak+1)/2d=-d/2d=-1/2(常数)
∴{1/(bn+1)}={1/(xk+1)}是以-1/2为公差的等差数列
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