已知数列{an}(n∈N*),首项a1=56,若二次方程anx2-an+1x-1=0的根α、β且满足3α+αβ+3β=1,则数列{an}的前n项和Sn=______.-数学

题目简介

已知数列{an}(n∈N*),首项a1=56,若二次方程anx2-an+1x-1=0的根α、β且满足3α+αβ+3β=1,则数列{an}的前n项和Sn=______.-数学

题目详情

已知数列{an}(n∈N*),首项a1=
5
6
,若二次方程anx2-an+1x-1=0的根α、β且满足3α+αβ+3β=1,则数列{an}的前n项和Sn=______.
题型:填空题难度:偏易来源:浦东新区二模

答案

依题意得:α+β=
an+1
an
,α•β=-class="stub"1
an

∵3α+αβ+3β=1,
∴3•
an+1
an
-class="stub"1
an
=1.
∴3an+1=an+1,
∴3(an+1-class="stub"1
2
)=an-class="stub"1
2

an+1-class="stub"1
2
an-class="stub"1
2
=class="stub"1
3
,又a1=class="stub"5
6

∴a1-class="stub"1
2
=class="stub"1
3

∴{an-class="stub"1
2
}是以class="stub"1
3
为首项,class="stub"1
3
为公比的等比数列.
∴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=a1+a2+…+an=[class="stub"1
3
+(class="stub"1
3
)
2
+…+(class="stub"1
3
)
n
]+class="stub"1
2
n
=class="stub"1
2
-class="stub"1
2
(class="stub"1
3
)
n
+class="stub"n
2

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

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