A. Maths新人..唔該幫幫

mathematical induction係一個用相對簡單的做法,證明對於命題,所有正整數皆成立的方法..

原理:

1.證明對於命題,在n=1時是成立

2.假設命題在n=k時成立, 證明它在n=k+1時也成立

3.得出結論

解釋: 當步驟1,2 做完以後,由於命題在n=1時成立,而當n=k=1成立,n=k+1=2亦成立,n= (k+1)+1 =3 也成立.......n= 4,5,6....78,79......100000000,199990000000000000....都成立>>> 並無限延伸...

amaths的數歸在算術不會太難,最重要是明白其原理.

其實還有另一種數歸,but佢係深d,唔係amaths課程度.

2008-07-14 11:45:30 補充：
有時未必由n=1做起,可能由n=2or n=3開始,按命題而定.

The purpose of MI is to prove a mathematical expression is true for all positive integer of n( that is why you see n in the expression).
Step 1 : To check that the expression is true for n=1, n=2 etc. If this is true, then
Step 2: Assume the expression is true for any value of n=k (that is the purpose of k). Then, the most important part of the problem is
Step 3: Prove that the expression is also true when n= k+1. That is when replacing k in step 2 by (k+1) is still true. That is when adding one more term to the expression of n=k you still come to the expression for (k+1).