等比数列{an}的首项为a1=2,公比q=3,则1a1a2+1a2a3+…+1anan+1=______.-数学

题目简介

等比数列{an}的首项为a1=2,公比q=3,则1a1a2+1a2a3+…+1anan+1=______.-数学

题目详情

等比数列{an}的首项为a1=2,公比q=3,则
1
a1a2
+
1
a2a3
+
+
1
anan+1
=______.
题型:填空题难度:中档来源:不详

答案

解;由题意可得,an=2•3n-1
class="stub"1
anan+1
=class="stub"1
4•32n-1
,则该数列是以class="stub"1
12
为首项,以class="stub"1
9
为公比的等比数列
class="stub"1
a1a2
+class="stub"1
a2a3
+
+class="stub"1
anan+1
=
class="stub"1
12
[1-(class="stub"1
9
)
n
]
1-class="stub"1
9
=class="stub"3
32
[1-(class="stub"1
9
)
n
]

故答案为:class="stub"3
32
[1-(class="stub"1
9
)
n
]

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