已知数列{an}中,a1=56,若以a1,a2,…,an为系数的二次方程an-1x2-anx+1=0(n∈N+,n≥2)都有根α,β且3α-αβ+3β=1,则{an}的前n项和Sn=______.-数

题目简介

已知数列{an}中,a1=56,若以a1,a2,…,an为系数的二次方程an-1x2-anx+1=0(n∈N+,n≥2)都有根α,β且3α-αβ+3β=1,则{an}的前n项和Sn=______.-数

题目详情

已知数列{an}中,a1=
5
6
,若以a1,a2,…,an为系数的二次方程an-1x2-anx+1=0(n∈N+,n≥2)都有根α,β且3α-αβ+3β=1,则{an}的前n项和Sn=______.
题型:填空题难度:中档来源:不详

答案

由题意,∵α+β=
an
an-1
,αβ=class="stub"1
an-1
代入3α-αβ+3β=1得an=class="stub"1
3
an-1+class="stub"1
3

an-class="stub"1
2
an-1-class="stub"1
2
=
class="stub"1
3
an-1+class="stub"1
3
-class="stub"1
2
an-1-class="stub"1
2
=class="stub"1
3
为定值.
∴数列{an-class="stub"1
2
}是等比数列.
∵a1-class="stub"1
2
=class="stub"5
6
-class="stub"1
2
=class="stub"1
3

∴an-class="stub"1
2
=class="stub"1
3
×( class="stub"1
3
)n-1=( class="stub"1
3
)n.
∴an=( class="stub"1
3
)n+class="stub"1
2

∴Sn=( class="stub"1
3
+class="stub"1
32
++class="stub"1
3n
)+class="stub"n
2
=
class="stub"1
3
(1-class="stub"1
3n
)
1-class="stub"1
3
+class="stub"n
2
=class="stub"n+1
2
-class="stub"1
3n

故答案为:Sn=class="stub"n+1
2
-class="stub"1
3n

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