(40) A.maths (Binomial Theorem) (2)
2007-10-30 6:41 am
Qs 22 only.
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Thanks for your help...
回答 (2)
2007-10-30 6:51 am
✔ 最佳答案
is the question" 1/(n!) - 1/[(n+1)!]?if yes,
1/(n!) - 1/[(n+1)!]?
= (n+1)/(n!)(n+1) - 1/[(n+1)!]
=(n+1)/[(n+1)!] - 1/[(n+1)!]
=(n+1-1)/[(n+1)!]
=n/[(n+1)!]
is it correct??i hope my answer can help you.
2007-10-30 6:50 am
1/n!﹣1/(n+1)!
= 1/n!﹣1/(n+1)n!
= (n+1)/(n+1)n!﹣1/(n+1)n!
= [(n+1)﹣1]/(n+1)n!
= (n+1﹣1)/(n+1)n!
= n/(n+1)n!
2007-10-29 22:54:31 補充：
最簡答案應該是= n/(n＋1)n!= n/(n＋1)n(n﹣1)!= 1/(n＋1)(n﹣1)!
= 1/n!﹣1/(n+1)n!
= (n+1)/(n+1)n!﹣1/(n+1)n!
= [(n+1)﹣1]/(n+1)n!
= (n+1﹣1)/(n+1)n!
= n/(n+1)n!
2007-10-29 22:54:31 補充：
最簡答案應該是= n/(n＋1)n!= n/(n＋1)n(n﹣1)!= 1/(n＋1)(n﹣1)!
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