问题 解答题
计算:
(1-
1
22
)×(1-
1
32
)
=______;
(1-
1
22
)×(1-
1
32
)×(1-
1
42
)
=______;
(1-
1
22
)×(1-
1
32
)×…×(1-
1
92
)×(1-
1
102
)
=______;
(1-
1
22
)×(1-
1
32
)×…×(1-
1
(n-1)2
)×(1-
1
n2
)
=______.
答案

(1-

1
22
)×(1-
1
32
)=
3
4
×
8
9
=
2
3

(1-

1
22
)×(1-
1
32
)×(1-
1
42
)=
2
3
×
15
16
=
5
8
=
4+1
2×4

(1-

1
22
)×(1-
1
32
)×(1-
1
42
)×(1-
1
52
)=
5
8
×
24
25
=
3
5
=
5+1
2×5

依此类推:(1-

1
22
)×(1-
1
32
)×…×(1-
1
92
)×(1-
1
102
)=
10+1
2×10
=
11
20

(1-

1
22
)×(1-
1
32
)×…×(1-
1
(n-1)2
)×(1-
1
n2
)=
n+1
2n

故答案为:

2
3
5
8
11
20
n+1
2n

问答题 简答题
名词解释