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
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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
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
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
是“ak*x^2+2a(k+1)*x+a(k+2)=0”吗?
①
ak*x^2+2a(k+1)*x+a(k+2)=0
a(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)]=0
d*x^2+2d*x+d=0
因d≠0
x^2+2x+1=0
(x+1)^2=0
x=-1(公共根)
证毕.
②
ak*x^2+2a(k+1)*x+a(k+2)=0
xk+(-1)=-2a(k+1)/ak
xk*(-1)=a(k+2)/ak
xk=-a(k+2)/ak
x1=-a3/a1=-(a1+2d)/a1=-1+2d/a1
x(k-1)=-a(k+1)/a(k-1)
1+xk=1-a(k+2)/ak=-2d/ak
1/[1+xk]=-ak/(2d)
所以
1/[1+x(k-1)]=-a(k-1)/(2d)
1/[1+xk]-1/[1+x(k-1)]=-ak/(2d)+a(k-1)/(2d)
=-[ak-a(k-1)]/(2d)
=-d/(2d)
=-1/2
所以1/(1+xn)是首项为-1+2d/a1,公差为-1/2的等差数列
①
ak*x^2+2a(k+1)*x+a(k+2)=0
a(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)]=0
d*x^2+2d*x+d=0
因d≠0
x^2+2x+1=0
(x+1)^2=0
x=-1(公共根)
证毕.
②
ak*x^2+2a(k+1)*x+a(k+2)=0
xk+(-1)=-2a(k+1)/ak
xk*(-1)=a(k+2)/ak
xk=-a(k+2)/ak
x1=-a3/a1=-(a1+2d)/a1=-1+2d/a1
x(k-1)=-a(k+1)/a(k-1)
1+xk=1-a(k+2)/ak=-2d/ak
1/[1+xk]=-ak/(2d)
所以
1/[1+x(k-1)]=-a(k-1)/(2d)
1/[1+xk]-1/[1+x(k-1)]=-ak/(2d)+a(k-1)/(2d)
=-[ak-a(k-1)]/(2d)
=-d/(2d)
=-1/2
所以1/(1+xn)是首项为-1+2d/a1,公差为-1/2的等差数列
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