標題:
關於分數問題
這條分數怎樣計？？ 1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72=?
最佳解答:
Please don't just post the steps ... You need to know the background information ... Back to the question , first , you need to learn these two formulae : 1. 1 / n( n + 1) = 1 / n - 1 / (n + 1) 2. d / n (n +1) (n + 2) = 1 / n - 1 / (n + d) Now , we can use the first formula to calculate : 1/2+1/6+1/12+1/20+1/ 30+1/42+1/56+1/72 = 1 / (1 x 2) + 1 /(2 x 3) + 1 / (3 x 4) + 1 / (4 x 5)+ ... + 1 / (8 x 9) =1 / 1 - 1 / 2 + 1 / 2 - 1 / 3 + 1 / 3 - 1 / 4 + 1 / 4 - 1 / 5 + 1 / 5 - ... + 1 / 8 - 1 / 9 = 1 - 1 / 9 = 8 / 9
其他解答:
1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72=8/9
關於分數問題
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發問:這條分數怎樣計？？ 1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72=?
最佳解答:
Please don't just post the steps ... You need to know the background information ... Back to the question , first , you need to learn these two formulae : 1. 1 / n( n + 1) = 1 / n - 1 / (n + 1) 2. d / n (n +1) (n + 2) = 1 / n - 1 / (n + d) Now , we can use the first formula to calculate : 1/2+1/6+1/12+1/20+1/ 30+1/42+1/56+1/72 = 1 / (1 x 2) + 1 /(2 x 3) + 1 / (3 x 4) + 1 / (4 x 5)+ ... + 1 / (8 x 9) =1 / 1 - 1 / 2 + 1 / 2 - 1 / 3 + 1 / 3 - 1 / 4 + 1 / 4 - 1 / 5 + 1 / 5 - ... + 1 / 8 - 1 / 9 = 1 - 1 / 9 = 8 / 9
其他解答:
1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72=8/9
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