關於Mathematical Induction

Let P(n)=1^2+2^2+3^2+...............+n^2 = n(n+1)(2n+1)/6
P(1)=1^2=1
RHS= (1)(1+1)(2(1)+1)/6=1
P(1) is true
Let P(k) is true
When n=k+1
1^2+2^2+3^2+...............+k^2+(k+1)^2 = k(k+1)(2k+1)/6+(k+1)^2
= (k+1)( k(2k+1)+6(k+1) ) /6
= (k+1)( 2k^2+k+6k+6)/6
= (k+1)( 2k^2+7k+6)/6
=(k+1)(2k+3)(k+2)/6
=(k+1)((k+1)+1)(2(k+1)+1)/6
If P(n) is true for n=k, it is also true for n=k+1. By the principle of mathematical induction, P is true for all positive integers n

背後原理其實你假設佢其中一個n是對的(即n是某一個正整數), 那麼佢下一個都岩, 所以1岩,2都岩, 直至所有正整數都岩.

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