讓我們並且考慮到紅色球的數量是一個正面奇數的案件, 並且在這種情況下讓T 是數字的產品在所有紅色球。讓G 是所有S 的可能的價值的總和(counting multiplicities) 並且H 是所有T 的可能的價值的總和(counting multiplicities) 。note that
G+H=(1+1/2)(1+1/3)...(1+1/1000)-1=(3/2)(4/3)...(1001/1000)-1=1001/2-1=999/2
since each possible value of S or T arises from a choice of whether each ball is red or blue, and the term −1 in the end eliminates the case where all balls are blue which is not allowed by the question. In a similar manner, we have
G-H=(1-1/2)1-1/3)...(1-1/1000)-1=(1/2)(2/3)...(999/1000)-1=-999/1000
since each possible value of T (corresponding to a choice of an odd number of red balls) is evaluated −1 time in the above product while each possible value of S is evaluated +1 time. It follows that
then,G=1/2(999/2-999/1000)=498501/2000
我要出街LOLZ..
諗下abc果題吧
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本帖最後由 『小龍龍﹏ 於 2008-3-2 11:13 編輯 ]