已知数列{an}的前n项和Sn满足Sn=2n+1,则当n≥2时,1a1+1a2+…+1an=______.-数学

题目简介

已知数列{an}的前n项和Sn满足Sn=2n+1,则当n≥2时,1a1+1a2+…+1an=______.-数学

题目详情

已知数列{an}的前n项和Sn满足Sn=2n+1,则当n≥2时,
1
a1
+
1
a2
+…+
1
an
=______.
题型:填空题难度:中档来源:重庆一模

答案

由Sn=2n+1 得,当n=1时,a1=S1=3,当n≥2时an=Sn-Sn-1=2 n-1∴an=
3   n=1
2n-1  n>1

class="stub"1
a1
+class="stub"1
a2
+…+class="stub"1
an
=class="stub"1
3
+class="stub"1
2
+class="stub"1
22
+…+class="stub"1
2n-1
=class="stub"1
3
+
class="stub"1
2
[1-(class="stub"1
2
)
n-1
]
1-class="stub"1
2
=class="stub"4
3
-(class="stub"1
2
)
n-1

故答案为:class="stub"4
3
-(class="stub"1
2
)
n-1

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