设Sn=12+16+112+…+1n(n+1),且Sn•Sn+1=34,则n的值为______.-数学

题目简介

设Sn=12+16+112+…+1n(n+1),且Sn•Sn+1=34,则n的值为______.-数学

题目详情

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

答案

由于class="stub"1
n(n+1)
=class="stub"1
n
-class="stub"1
n+1

Sn=class="stub"1
2
+class="stub"1
6
+…+class="stub"1
n(n+1)
=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

SnSn+1=class="stub"n
n+1
•class="stub"n+1
n+2
=class="stub"n
n+2
=class="stub"3
4

∴n=6
故答案为:6

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