问题 填空题
设f(n)=
1
n+1
+
2
n+2
+
1
n+3
+…+
1
2n
(n∈N*)
,那么f(n+1)-f(n)=
1
n+2
+
1
n+3
+…+
1
2n
+
1
2n+1
+
1
2n+2
-(
1
n+1
+
1
n+2
+…+
1
2n
)
=
1
2n+1
+
1
2n+2
-
1
n+1
=______.
答案

∵f(n)=

1
n+1
+
2
n+2
+
1
n+3
+…+
1
2n
(n∈N*),、

∴f(n+1)=

1
n+2
+
1
n+3
+…+
1
2n
+
1
2n+1
+
1
2n+2

∴f(n+1)-f(n)=

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

=

1
2n+1
+
1
2n+2
-
1
n+1

=

1
2n+1
-
1
2n+2

故答案为:

1
2n+1
-
1
2n+2

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