> 数学 >
计算：(1−
1
22
)(1−
1
32
)(1−
1
42
)…(1−
1
20042
)(1−
1
20052
)
人气：268 ℃　时间：2020-01-30 13:29:51
解答
(1−
1
22
)(1−
1
32
)(1−
1
42
)…(1−
1
20042
)(1−
1
20052
)=
22−1
22
•
32−1
32
•
42−1
42
…
20052−1
20052
=
1
2
×
2006
2005
=
1003
2005
．
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