试题:
1+
1
12
+
1
22
=______;
1+
1
12
+
1
22
+
1+
1
22
+
1
32
=______;
1+
1
12
+
1
22
+
1+
1
22
+
1
32
+
1+
1
32
+
1
42

=______;由此猜想
1+
1
n2
+
1
(n+1)2
=______;
1+
1
12
+
1
22
+
1+
1
22
+
1
32
+
1+
1
32
+
1
42
+…+
1+
1
20032
+
1
20042
=______.

答案:

我来补答
1+
1
12
+
1
22
=
9
4
=
3
2
=1
1
2

1+
1
12
+
1
22
+
1+
1
22
+
1
32
=
3
2
+
49
36
=
3
2
+
7
6
=
16
6
=
8
3
=2
2
3

1+
1
12
+
1
22
+
1+
1
22
+
1
32
+
1+
1
32
+
1
42
=
3
2
+
7
6
+
13
12
=
15
4
=3
3
4

猜想:
1+
1
n2
+
1
(n+1)2
=1+
1
n(n+1)

1+
1
12
+
1
22
+
1+
1
22
+
1
32
+
1+
1
32
+
1
42
+…+
1+
1
20032
+
1
20042

=
3
2
+
7
6
+
13
12
+…+(1-
1
2003×2004
),
=(1+1-
1
2
)+(1+
1
2
-
1
3
)+(1+
1
3
-
1
4
)+…+(1+
1
2003
-
1
2004
),
=2003+1-
1
2004

=2003
2003
2004

故答案为:1
1
2
;2
2
3
;3
3
4
;1+
1
n(n+1)
;2003
2003
2004
 
 
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