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已知2+23=22×23,3+38=32×38,4+415=42×415,…;请你观察后,找出规律,并写出一组等式5+524=52×5245+524=52×524,若用n(n为正整数)表示上面的规律为(n+1)+n+1(n+1)2−1=(n+1)2•n+1(n+1)2−1(n+1)+n
题目详情
已知2+
=22×
,3+
=32×
,4+
=42×
,…;请你观察后,找出规律,并写出一组等式
=22×
,3+
=32×
,4+
=42×
,…;请你观察后,找出规律,并写出一组等式
2 2 3 3 22×
,3+
=32×
,4+
=42×
,…;请你观察后,找出规律,并写出一组等式
,3+
=32×
,4+
=42×
,…;请你观察后,找出规律,并写出一组等式
,3+
=32×
,4+
=42×
,…;请你观察后,找出规律,并写出一组等式
2 2 3 3
3 3 8 8 32×
,4+
=42×
,…;请你观察后,找出规律,并写出一组等式
,4+
=42×
,…;请你观察后,找出规律,并写出一组等式
,4+
=42×
,…;请你观察后,找出规律,并写出一组等式
3 3 8 8
4 4 15 15 42×
,…;请你观察后,找出规律,并写出一组等式
,…;请你观察后,找出规律,并写出一组等式
,…;请你观察后,找出规律,并写出一组等式
4 4 15 15
=52×
5 5 24 24 52×
52×
2×
5 5 24 24
=52×
5 5 24 24 52×
52×
2×
5 5 24 24
=(n+1)2•
n+1 n+1 (n+1)2−1 (n+1)2−1 (n+1)2−1(n+1)2−12−1(n+1)2•
(n+1)2•
2•
n+1 n+1 (n+1)2−1 (n+1)2−1 (n+1)2−1(n+1)2−12−1
=(n+1)2•
n+1 n+1 (n+1)2−1 (n+1)2−1 (n+1)2−1(n+1)2−12−1(n+1)2•
(n+1)2•
2•
n+1 n+1 (n+1)2−1 (n+1)2−1 (n+1)2−1(n+1)2−12−1
2 |
3 |
2 |
3 |
3 |
8 |
3 |
8 |
4 |
15 |
4 |
15 |
5+
=52×
5 |
24 |
5 |
24 |
5+
=52×
,若用n(n为正整数)表示上面的规律为5 |
24 |
5 |
24 |
(n+1)+
=(n+1)2•
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
(n+1)+
=(n+1)2•
.2+n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
2 |
3 |
2 |
3 |
3 |
8 |
3 |
8 |
4 |
15 |
4 |
15 |
5+
=52×
5 |
24 |
5 |
24 |
5+
=52×
,若用n(n为正整数)表示上面的规律为5 |
24 |
5 |
24 |
(n+1)+
=(n+1)2•
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
(n+1)+
=(n+1)2•
.n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
2 |
3 |
2 |
3 |
3 |
8 |
3 |
8 |
4 |
15 |
4 |
15 |
5+
=52×
5 |
24 |
5 |
24 |
5+
=52×
,若用n(n为正整数)表示上面的规律为5 |
24 |
5 |
24 |
(n+1)+
=(n+1)2•
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
(n+1)+
=(n+1)2•
.22×n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
2 |
3 |
3 |
8 |
3 |
8 |
4 |
15 |
4 |
15 |
5+
=52×
5 |
24 |
5 |
24 |
5+
=52×
,若用n(n为正整数)表示上面的规律为5 |
24 |
5 |
24 |
(n+1)+
=(n+1)2•
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
(n+1)+
=(n+1)2•
.2×n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
2 |
3 |
3 |
8 |
3 |
8 |
4 |
15 |
4 |
15 |
5+
=52×
5 |
24 |
5 |
24 |
5+
=52×
,若用n(n为正整数)表示上面的规律为5 |
24 |
5 |
24 |
(n+1)+
=(n+1)2•
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
(n+1)+
=(n+1)2•
.n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
2 |
3 |
3 |
8 |
3 |
8 |
4 |
15 |
4 |
15 |
5+
=52×
5 |
24 |
5 |
24 |
5+
=52×
,若用n(n为正整数)表示上面的规律为5 |
24 |
5 |
24 |
(n+1)+
=(n+1)2•
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
(n+1)+
=(n+1)2•
.32×n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
3 |
8 |
4 |
15 |
4 |
15 |
5+
=52×
5 |
24 |
5 |
24 |
5+
=52×
,若用n(n为正整数)表示上面的规律为5 |
24 |
5 |
24 |
(n+1)+
=(n+1)2•
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
(n+1)+
=(n+1)2•
.2×n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
3 |
8 |
4 |
15 |
4 |
15 |
5+
=52×
5 |
24 |
5 |
24 |
5+
=52×
,若用n(n为正整数)表示上面的规律为5 |
24 |
5 |
24 |
(n+1)+
=(n+1)2•
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
(n+1)+
=(n+1)2•
.n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
3 |
8 |
4 |
15 |
4 |
15 |
5+
=52×
5 |
24 |
5 |
24 |
5+
=52×
,若用n(n为正整数)表示上面的规律为5 |
24 |
5 |
24 |
(n+1)+
=(n+1)2•
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
(n+1)+
=(n+1)2•
.42×n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
4 |
15 |
5+
=52×
5 |
24 |
5 |
24 |
5+
=52×
,若用n(n为正整数)表示上面的规律为5 |
24 |
5 |
24 |
(n+1)+
=(n+1)2•
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
(n+1)+
=(n+1)2•
.2×n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
4 |
15 |
5+
=52×
5 |
24 |
5 |
24 |
5+
=52×
,若用n(n为正整数)表示上面的规律为5 |
24 |
5 |
24 |
(n+1)+
=(n+1)2•
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
(n+1)+
=(n+1)2•
.n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
4 |
15 |
5+
=52×
5+5 |
24 |
5 |
24 |
5 |
24 |
5 |
24 |
5 |
24 |
5 |
24 |
5 |
24 |
5 |
24 |
5 |
24 |
5+
=52×
5+5 |
24 |
5 |
24 |
5 |
24 |
5 |
24 |
5 |
24 |
5 |
24 |
5 |
24 |
5 |
24 |
5 |
24 |
(n+1)+
=(n+1)2•
(n+1)+n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
(n+1)+
=(n+1)2•
(n+1)+n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
n+1 |
(n+1)2−1 |
▼优质解答
答案和解析
∵2+23=22×23,3+38=32×38,4+415=42×415,所以写一组等式为 5+524=52×524,若用n(n为正整数)表示上面的规律为 (n+1)+n+1(n+1)2−1=(n+1)2•n+1(n+1)2−1.故答案为:5+524=52×524,(n+1)+n+1(n+1)2−1...
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