设数列{an}满足:a1=56,且以a1,a2,a3,…,an为系数的一元二次方程an-1x2-anx+1=0(n∈N*,n≥2)都有根α,β,且两个根α,β满足3α-αβ+3β=1.(1)求数列{a

题目简介

设数列{an}满足:a1=56,且以a1,a2,a3,…,an为系数的一元二次方程an-1x2-anx+1=0(n∈N*,n≥2)都有根α,β,且两个根α,β满足3α-αβ+3β=1.(1)求数列{a

题目详情

设数列{an}满足:a1=
5
6
,且以a1,a2,a3,…,an为系数的一元二次方程an-1x2-anx+1=0(n∈N*,n≥2)都有根α,β,且两个根α,β满足3α-αβ+3β=1.
(1)求数列{an}的通项an
(2)求{an}的前n项和Sn
题型:解答题难度:中档来源:不详

答案

由3α-αβ+3β=1及韦达定理得3(α+β)-αβ=3
an
an-1
-class="stub"1
an-1
=1⇒
an=class="stub"1
3
an-1+class="stub"1
3
(n∈N*,n≥2)

(1)设有λ满足an+λ=class="stub"1
3
(an-1+λ)⇒λ=-class="stub"1
2
,即an-class="stub"1
2
=class="stub"1
3
(an-1-class="stub"1
2
)

所以数列{an-class="stub"1
2
]是以(a1-class="stub"1
2
)为首项,class="stub"1
3
为公比的等比数列.
所以an-class="stub"1
2
=(a1-class="stub"1
2
)•(class="stub"1
3
)n-1
an=(class="stub"1
3
)n+class="stub"1
2
(n∈N*)
(2)Sn=a1+a2++an=class="stub"1
3
+(class="stub"1
3
)2++(class="stub"1
3
)n+class="stub"n
2
=class="stub"n+1
2
-class="stub"1
2
•(class="stub"1
3
)n

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