设Sn=12+16+112+…+1n(n+1)(n∈N*),且Sn+1•Sn+2=34,则n的值是______.-数学

题目简介

设Sn=12+16+112+…+1n(n+1)(n∈N*),且Sn+1•Sn+2=34,则n的值是______.-数学

题目详情

Sn=
1
2
+
1
6
+
1
12
+…+
1
n(n+1)
(n∈N*),且Sn+1Sn+2=
3
4
,则n的值是______.
题型:填空题难度:中档来源:不详

答案

class="stub"1
n(n+1)
=class="stub"1
n
-class="stub"1
n+1
,∴Sn=(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

∴S n+1•Sn+2=class="stub"n+1
n+2
•class="stub"n+2
n+3
=class="stub"n+1
n+3
=class="stub"3
4
,解得n=5
故答案为:5.

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