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標題: [AMaths] 終極MI(已修正) [打印本頁]

作者: bc123456    時間: 2010-5-3 09:13 AM     標題: 終極MI(已修正)

唔洗多講, 睇題目

Prove, by mathematical induction, that
(n+1)(n+2)(n+3)...(2n)=(2^n)×1×3×5×...×(2n-1)
for all positive integers n.

solution: (by asdfzxc806)
when n=1
LHS=2
RHS=(2)(1)=2
P(1) is true
when n=k+1
LHS=(k+2)(k+3)(k+4)...(2k)(2k+1)(2k+2)
     =(2^k)×1×3×5×...×(2k-1)(2k+1)(2k+2)/(k+1)
     =2^(k+1)x1x3x5x...x(2k+1)
RHS=2^(k+1)x1x3x5x...x(2k+1)
P(k+1) is true

另一題MI: (by桃子)
Given that a and b are positive integers. By mathematical induction, show that
(a+b)^(2n+1)-a^(2n+1)-b^(2n+1) is divisible by ab(a+b) for all positive integers n.

If it looks too difficult, you may first try the special case a=2, b=3.
52n+1-22n+1-32n+1 is divisible by 30 for all positive integers na
附上個人見解:
when n=1
P(n)=(a+b)^3-a^3-b^3
   =3a^2b+3ab^2
     =3ab(a+b)   which is divisible by ab(a+b)
Assume that P(k) is true
i.e. (a+b)^(2k+1)-a^(2k+1)-b^(2k+1)=Mab(a+b)     where M is an integer
跟住見下圖
最後抽common factor


[ 本帖最後由 bc123456 於 2010-5-3 12:26 PM 編輯 ]
作者: skysiu15    時間: 2010-5-3 09:27 AM

真係好終極....
我a-maths太差~唔識做
樓主有無ans?
作者: bc123456    時間: 2010-5-3 09:30 AM

有, 遲d先post
作者: ktp11364    時間: 2010-5-3 10:33 AM

左邊係quadratic equation,右邊係linear equation wo...
L.H.S. = (2^k)(2k-1)(2k+2)
R.H.S. = (2^(k+1))(2k+1)
小弟不才,請賜教
作者: death_kg    時間: 2010-5-3 10:38 AM

見過幾次啦 唔算終極
仲要好似最近先有人問過
作者: gameson    時間: 2010-5-3 10:40 AM

咁都叫終極? 不是吧
作者: asdfzxc806    時間: 2010-5-3 10:43 AM

引用:
原帖由 bc123456 於 2010-5-3 09:13 AM 發表
唔洗多講, 睇題目

Prove, by mathematical induction, that
(n+1)(n+2)(n+3)...(2n)=(2^n)×1×3×5×...×(2n-1)
for all positive integers n.
when n=1
LHS=2
RHS=(2)(1)=2
P(1) is true
when n=k+1
LHS=(k+2)(k+3)(k+4)...(2k)(2k+1)(2k+2)
     =(2^k)×1×3×5×...×(2k-1)(2k+1)(2k+2)/(k+1)
     =2^(k+1)x1x3x5x...x(2k+1)
RHS=2^(k+1)x1x3x5x...x(2k+1)
P(k+1) is true
其實呢題都唔算難...仲有d係仲難d既...
作者: skysiu15    時間: 2010-5-3 11:00 AM

小弟不才..想問下點解RHS=(2)(1)=2?
因為RHS係=(2^n)×1×3×5×...×(2n-1)...如果n=1...RHS點解會係2?...後面X3X5X....唔洗理?定係點?
作者: sport    時間: 2010-5-3 11:08 AM

when n=1, RHS=(2^1)×(2*1-1)=2
引用:
原帖由 asdfzxc806 於 2010-5-3 10:43 AM 發表


LHS=(k+2)(k+3)(k+4)...(2k)(2k+1)(2k+2)
     =(2^k)×1×3×5×...×(2k-1)(2k+1)(2k+2)/(k+1)
     
How can you prove???
作者: 桃子    時間: 2010-5-3 11:10 AM

Try this one.
Given that a and b are positive integers. By mathematical induction, show that
(a+b)[suptag]2n+1[/suptag]-a[suptag]2n+1[/suptag]-b[suptag]2n+1[/suptag] is divisible by ab(a+b) for all positive integers n.

If it looks too difficult, you may first try the special case a=2, b=3.
5[suptag]2n+1[/suptag]-2[suptag]2n+1[/suptag]-3[suptag]2n+1[/suptag] is divisible by 30 for all positive integers n
(a question that appeared in this forum a few times)
作者: felix139139139    時間: 2010-5-3 11:11 AM

引用:
原帖由 asdfzxc806 於 2010-5-3 10:43 AM 發表

when n=1
LHS=2
RHS=(2)(1)=2
P(1) is true
when n=k+1
LHS=(k+2)(k+3)(k+4)...(2k)(2k+1)(2k+2)
     =(2^k)×1×3×5×...×(2k-1)(2k+1)(2k+2)/(k+1)
    =2^(k+1)x1x3x5x...x(2k+1)
RHS=2^(k+1) ...
呢步你點跳?
(2^k)x1x3x5x...(2k-1)(2k+1)(2k+2)/(k+1)
=(2^k)x1x3x5...(2k-1)(2k+1)2(k+1)/(k+1)
我計極都係
=(2^k+1)x1x3x5x...(2k-1)(2k+1)


[ 本帖最後由 felix139139139 於 2010-5-3 11:40 AM 編輯 ]
作者: lopo42    時間: 2010-5-3 11:12 AM     標題: 回覆 #9 sport 的帖子

LHS=(k+2)(k+3)(k+4)...(2k)(2k+1)(2k+2)
     =(k+1)(k+2)(k+3)(k+4)...(2k)(2k+1)(2k+2)/(k+1)
     =(2^k)×1×3×5×...×(2k-1)(2k+1)(2k+2)/(k+1)         (assumption,(k+1)(k+2)(k+3)(k+4)...(2k)=2[suptag]k[/suptag]×1×3×5×...×(2k-1) )

[ 本帖最後由 lopo42 於 2010-5-3 11:15 AM 編輯 ]
作者: lopo42    時間: 2010-5-3 11:17 AM     標題: 回覆 #11 felix139139139 的帖子

我諗你寫(2^k+1)係唔係指2[suptag]k+1[/suptag],如果係就岩
作者: asdfzxc806    時間: 2010-5-3 11:21 AM

引用:
原帖由 felix139139139 於 2010-5-3 11:11 AM 發表


呢步你點跳?
(2^k)x1x3x5x...(2k-1)(2k+1)(2k+2)/(k+1)
=(2^k)x1x3x5...(2k-1)(2k+1)2(k+1)/(k+1)
我計極都係
=(2^k+1)x1x3x5x...(2k-1)(2k+1)  
兩耆一樣嫁....
就好似
1x3x5x7x9x...x21=1x3x5x7x9x...x19x21
再一個example
1x3x5x...(2n+1)=1x3x5x...(2n-5)(2n-3)(2n-1)(2n+1)
我只係將1d無關既term"隱藏"左係...入面
引用:
原帖由 桃子 於 2010-5-3 11:10 AM 發表
Try this one.
Given that a and b are positive integers. By mathematical induction, show that
(a+b)2n+1-a2n+1-b2n+1 is divisible by ab(a+b) for all positive integers n.

If it looks too diff ...
見到條數咁長我都唔想睇=.=
過左RS我先睇啦
作者: felix139139139    時間: 2010-5-3 11:39 AM

引用:
原帖由 asdfzxc806 於 2010-5-3 11:21 AM 發表

兩耆一樣嫁....
就好似
1x3x5x7x9x...x21=1x3x5x7x9x...x19x21
再一個example
1x3x5x...(2n+1)=1x3x5x...(2n-5)(2n-3)(2n-1)(2n+1)
我只係將1d無關既term"隱藏"左係...入面

見到條數咁長我都唔想睇=. ...
係wor...
計到個腦實晒,諗唔到野
作者: bc123456    時間: 2010-5-3 11:59 AM

push...睇下我做得畯
作者: lopo42    時間: 2010-5-3 12:07 PM

引用:
原帖由 bc123456 於 2010-5-3 11:59 AM 發表
push...睇下我做得飱唔飱
尾三點去尾二?
作者: bc123456    時間: 2010-5-3 12:12 PM

引用:
原帖由 lopo42 於 2010-5-3 12:07 PM 發表


尾三點去尾二?
睇錯左tim...改番...

[ 本帖最後由 bc123456 於 2010-5-3 01:33 PM 編輯 ]
作者: lopo42    時間: 2010-5-3 01:56 PM

引用:
原帖由 桃子 於 2010-5-3 11:10 AM 發表
Try this one.
Given that a and b are positive integers. By mathematical induction, show that
(a+b)2n+1-a2n+1-b2n+1 is divisible by ab(a+b) for all positive integers n.

If it looks too diff ...
alternative method
作者: 桃子    時間: 2010-5-3 02:01 PM

引用:
原帖由 lopo42 於 2010-5-3 01:56 PM 發表

alternative method
It works.

By the way, this can be solved much easier by modular arithmetic.

Using mathematical induction is really really really tedious.
作者: bc123456    時間: 2010-5-3 02:06 PM

兩位數學知識咁豐富, 我淨係識用CE level既方法
作者: lopo42    時間: 2010-5-3 02:09 PM

引用:
原帖由 bc123456 於 2010-5-3 02:06 PM 發表
兩位數學知識咁豐富, 我淨係識用CE level既方法
過獎....
相信桃子的數學知識比我豐富更多
bc123456 , 加油




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