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[AMaths] 終極MI(已修正)

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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 ]
附件: 您所在的用戶組無法下載或查看附件
   

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真係好終極....
我a-maths太差~唔識做
樓主有無ans?

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有, 遲d先post

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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

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見過幾次啦 唔算終極
仲要好似最近先有人問過

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咁都叫終極? 不是吧

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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...

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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?

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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???

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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 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 n
(a question that appeared in this forum a few times)

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