数列{an}中,a2=2,an,an+1是方程x2-(2n+1)x+1bn=0的两个根,则数列{bn}的前n项和Sn=______.-高二数学

题目简介

数列{an}中,a2=2,an,an+1是方程x2-(2n+1)x+1bn=0的两个根,则数列{bn}的前n项和Sn=______.-高二数学

题目详情

数列{an}中,a2=2,an,an+1是方程x2-(2n+1)x+
1
bn
=0
的两个根,则数列{bn}的前n项和Sn=______.
题型:填空题难度:中档来源:不详

答案

∵an,an+1是方程x2-(2n+1)x+class="stub"1
bn
=0
的两个根,
∴an+an+1=2n+1,anan+1=class="stub"1
bn

∵a2=2,∴a1=2+1-2=1,
∴an-n=-[an+1-(n+1)],
∴an=n
anan+1=class="stub"1
bn

∴bn=class="stub"1
n(n+1)
=class="stub"1
n
-class="stub"1
n+1

∴Sn=b1+b2+…+bn
=(1-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+…+(class="stub"1
n
-class="stub"1
n+1

=1-class="stub"1
n+1

=class="stub"n
n+1

故答案为:class="stub"n
n+1

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