设二次方程anx^2-a(n+1)x+1=0有两个根x1,x2,且满足6x1-2x1x2+6x2=3.且a1=1设二次方程anx²-a(n+1)x+1=0(n=1,2,3…)有两根α和β,且满足6α-2αβ+6β=3。,a1=1(1)试用an表示a(n+1);(2)求证:{an-2/3}是等比数
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![设二次方程anx^2-a(n+1)x+1=0有两个根x1,x2,且满足6x1-2x1x2+6x2=3.且a1=1设二次方程anx²-a(n+1)x+1=0(n=1,2,3…)有两根α和β,且满足6α-2αβ+6β=3。,a1=1(1)试用an表示a(n+1);(2)求证:{an-2/3}是等比数](/uploads/image/z/980875-19-5.jpg?t=%E8%AE%BE%E4%BA%8C%E6%AC%A1%E6%96%B9%E7%A8%8Banx%5E2-a%28n%2B1%29x%2B1%3D0%E6%9C%89%E4%B8%A4%E4%B8%AA%E6%A0%B9x1%2Cx2%2C%E4%B8%94%E6%BB%A1%E8%B6%B36x1-2x1x2%2B6x2%3D3.%E4%B8%94a1%3D1%E8%AE%BE%E4%BA%8C%E6%AC%A1%E6%96%B9%E7%A8%8Banx%26sup2%3B-a%28n%2B1%29x%2B1%3D0%28n%3D1%2C2%2C3%E2%80%A6%29%E6%9C%89%E4%B8%A4%E6%A0%B9%CE%B1%E5%92%8C%CE%B2%EF%BC%8C%E4%B8%94%E6%BB%A1%E8%B6%B36%CE%B1-2%CE%B1%CE%B2%2B6%CE%B2%3D3%E3%80%82%2Ca1%3D1%EF%BC%881%EF%BC%89%E8%AF%95%E7%94%A8an%E8%A1%A8%E7%A4%BAa%EF%BC%88n%2B1%29%EF%BC%9B%EF%BC%882%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9A%7Ban-2%2F3%7D%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0)
设二次方程anx^2-a(n+1)x+1=0有两个根x1,x2,且满足6x1-2x1x2+6x2=3.且a1=1设二次方程anx²-a(n+1)x+1=0(n=1,2,3…)有两根α和β,且满足6α-2αβ+6β=3。,a1=1(1)试用an表示a(n+1);(2)求证:{an-2/3}是等比数
设二次方程anx^2-a(n+1)x+1=0有两个根x1,x2,且满足6x1-2x1x2+6x2=3.且a1=1
设二次方程anx²-a(n+1)x+1=0(n=1,2,3…)有两根α和β,且满足6α-2αβ+6β=3。,a1=1
(1)试用an表示a(n+1);
(2)求证:{an-2/3}是等比数列;
(3)求数列{an}的通项公式。
设二次方程anx^2-a(n+1)x+1=0有两个根x1,x2,且满足6x1-2x1x2+6x2=3.且a1=1设二次方程anx²-a(n+1)x+1=0(n=1,2,3…)有两根α和β,且满足6α-2αβ+6β=3。,a1=1(1)试用an表示a(n+1);(2)求证:{an-2/3}是等比数
6α-2αβ+6β=3
6(α+β)-2αβ=3
6a(n+1)/an -2/an=3
a(n+1)=(1/2)an+(1/3)
a(n+1)-(2/3)=(1/2)an+(1/3)-(2/3)=(1/2)[an-(2/3)]
所以:{an-2/3}是公比为1/2的等比数列
设bn=an-(2/3)
则:b1=a1-(2/3)=1/3
bn=b1*(1/2)^(n-1)=(1/3)*(1/2)^(n-1)=(2/3)*2^(-n)
an-(2/3)=(2/3)*2^(-n)
an=(2/3)+(2/3)*2^(-n)