已知数列{an},an=2n,则1a1+1a2+…+1an=______.-数学

题目简介

已知数列{an},an=2n,则1a1+1a2+…+1an=______.-数学

题目详情

已知数列{an},an=2n,则
1
a1
+
1
a2
+…+
1
an
=______.
题型:填空题难度:中档来源:不详

答案

由题意得:数列{an}为首项是2,公比为2的等比数列,
由an=2n,得到数列{an}各项为:2,22,…,2n,
class="stub"1
a1
+class="stub"1
a2
+…+class="stub"1
an
=class="stub"1
2
+class="stub"1
22
+…+class="stub"1
2n

∴数列{class="stub"1
an
}是首项为class="stub"1
2
,公比为class="stub"1
2
的等比数列,
class="stub"1
a1
+class="stub"1
a2
+…+class="stub"1
an
=class="stub"1
2
+class="stub"1
22
+…+class="stub"1
2n
=
class="stub"1
2
[1-(class="stub"1
2
)
n
]
1-class="stub"1
2
=1-class="stub"1
2n

故答案为:1-class="stub"1
2n

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