问题 填空题
Sn=
1
2
+
1
6
+
1
12
+…+
1
n(n+1)
, 且 SnSn+1=
3
4
,则n的值为______.
答案

由于

1
n(n+1)
=
1
n
-
1
n+1

Sn=

1
2
+
1
6
+…+
1
n(n+1)
=1-
1
2
+
1
2
-
1
3
+…+
1
n
-
1
n+1

=1-

1
n+1
=
n
n+1

SnSn+1=

n
n+1
n+1
n+2
=
n
n+2
=
3
4

∴n=6

故答案为:6

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