已知关于x的二次方程an x2-a(n+1) x+1=0(n∈N*)的两根α,β满足6α-2αβ+6β=3,且a1=1(1)求数列的通项公式an;(2)求数列an的前n项和Sn
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![已知关于x的二次方程an x2-a(n+1) x+1=0(n∈N*)的两根α,β满足6α-2αβ+6β=3,且a1=1(1)求数列的通项公式an;(2)求数列an的前n项和Sn](/uploads/image/z/10184129-17-9.jpg?t=%E5%B7%B2%E7%9F%A5%E5%85%B3%E4%BA%8Ex%E7%9A%84%E4%BA%8C%E6%AC%A1%E6%96%B9%E7%A8%8Ban+x2-a%28n%2B1%29+x%2B1%3D0%EF%BC%88n%E2%88%88N%2A%EF%BC%89%E7%9A%84%E4%B8%A4%E6%A0%B9%CE%B1%2C%CE%B2%E6%BB%A1%E8%B6%B36%CE%B1-2%CE%B1%CE%B2%2B6%CE%B2%3D3%2C%E4%B8%94a1%3D1%281%29%E6%B1%82%E6%95%B0%E5%88%97%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8Fan%EF%BC%9B%282%29%E6%B1%82%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn)
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已知关于x的二次方程an x2-a(n+1) x+1=0(n∈N*)的两根α,β满足6α-2αβ+6β=3,且a1=1(1)求数列的通项公式an;(2)求数列an的前n项和Sn
已知关于x的二次方程an x2-a(n+1) x+1=0(n∈N*)的两根α,β满足6α-2αβ+6β=3,且a1=1
(1)求数列的通项公式an;(2)求数列an的前n项和Sn
已知关于x的二次方程an x2-a(n+1) x+1=0(n∈N*)的两根α,β满足6α-2αβ+6β=3,且a1=1(1)求数列的通项公式an;(2)求数列an的前n项和Sn
α+β=a(n+1)/an,αβ=1/an
6α-2αβ+6β=3
6(α+β)-2αβ=3
6a(n+1)/an-2/an=3
a(n+1)-2/3=1/2(an-2/3)
[a(n+1)-2/3]/(an-2/3)=1/2
an-2/3=(a1-2/3)(1/2)^(n-1),a1=1
an=1/3*(1/2)^(n-1)+2/3
当n=1时a1=1,满足a1=1
an=1/3*(1/2)^(n-1)+2/3
2)Sn=1/3[(1+(1/2)+(1/2)^2+...+(1/2)^(n-1)]+2n/3
=2/3[1-(1/2)^(n-1)]+2n/3