今天我睡午覺睡睡下突然想通以下的問題！ｈａｈａ 大家不防研究一下！
Special MI
Question: The product of any n consecutive integers is divisible by n!
Answer: We can change it to mathematical statement
P(x,y): x(x+1)(x+2)...(x+(y-1))=My! <--y consective integer start from x
P(x,1): x=x*1! which is obvious true any positive integer x
So P(1,1)->P(2,1)->P(3,1)->...->P(x,1)->...
P(1,y): 1*(1+1)*(1+2)...(1+(y-1))=1*y! which is obvious true any positive integer y
So P(1,1)->P(1,2)->P(1,3)->...->P(1,y)->...
Assume P(a-1,b+1) is true, (a-1)(a-1+1)...[(a-1)+(b+1)-1]=M(b+1)! {a>=2,b>=1}
and P(a,b) is true, (a)(a+1)...[a+(b+1-1)]=N(b!)
P(a,b):
LHS=a(a+1)(a+2)...(a+b-1)[a+(b+1)-1]
=a(a+1)(a+2)...(a+b-1)[(a-1)+(b+1)]
=[a(a+1)(a+2)...(a+b-1)(a-1)]+[a(a+1)(a+2)...(a+b-1)(b+1)]
=[(a-1)a(a+1)(a+2)...(a+b-1)]+[a(a+1)(a+2)...(a+b-1)(b+1)]
=M(b+1)!+N(b!)(b+1)
=[M+N](b+1)!
=RHS
So P(a,b) is true. By the principle of MI, P(a,b) is true for any positive integers a,b.
Remarks: Why it works? Think it carefully, I will discuss with you all when we meet!
你們可能會問，阿ｓｉｒ日記也打數學是不是瘋了？我答你．．．是的！我是瘋狂沉迷，不比那些沉迷打機，賭博，愛情的人好了多小 ！
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