1+112+122=______;1+112+122+1+122+132=______;1+112+122+1+122+132+1+132+142=______;由此猜想1+1n2+1(n+1)2=_

题目简介

1+112+122=______;1+112+122+1+122+132=______;1+112+122+1+122+132+1+132+142=______;由此猜想1+1n2+1(n+1)2=_

题目详情

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
=______.
题型:解答题难度:中档来源:不详

答案

1+class="stub"1
12
+class="stub"1
22
=
class="stub"9
4
=class="stub"3
2
=1class="stub"1
2

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

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

猜想:
1+class="stub"1
n2
+class="stub"1
(n+1)2
=1+class="stub"1
n(n+1)

1+class="stub"1
12
+class="stub"1
22
+
1+class="stub"1
22
+class="stub"1
32
+
1+class="stub"1
32
+class="stub"1
42
+…+
1+class="stub"1
20032
+class="stub"1
20042

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

=2003class="stub"2003
2004

故答案为:1class="stub"1
2
;2class="stub"2
3
;3class="stub"3
4
;1+class="stub"1
n(n+1)
;2003class="stub"2003
2004

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