Quod Erat Demonstrandum

2010/09/28

唔用 MI

Filed under: Pure Mathematics — johnmayhk @ 12:36 下午

以下例題在 mathematical induction 一課:

Prove that

1. Show that the product of 4 consecutive positive integers is divisible by 4!.

2. Show that 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{n}} > 2(\sqrt{n + 1} - 1) (\forall n \in \mathbb{N}).

但,它們不一定以 M.I.處理。

1. 對任何正整數 n,因

C_4^{n+3} = \frac{n(n+1)(n+2)(n+3)}{4!}

C_4^{n+3} 是正整數,故 4! 整除 n(n+1)(n+2)(n+3)。證畢。

其實,更一般的結果是:

“The product of r consecutive integers (注:不一定是 positive integers) is divisible by r!"

2. 如何得出那個不等式?不過是節節相消,由 \sqrt{k+1} - \sqrt{k} 出發。

對正整數 k,考慮

\sqrt{k+1} - \sqrt{k}
= (\sqrt{k+1} - \sqrt{k})(\frac{\sqrt{k+1} + \sqrt{k}}{\sqrt{k+1} + \sqrt{k}})
= \frac{1}{\sqrt{k+1} + \sqrt{k}}
< \frac{1}{\sqrt{k} + \sqrt{k}}
= \frac{1}{2\sqrt{k}}

2(\sqrt{k+1} - \sqrt{k}) < \frac{1}{\sqrt{k}}

好了,加個 summation sign,以求節節相消是也,

2\displaystyle \sum_{k=1}^{n} (\sqrt{k+1} - \sqrt{k}) < \displaystyle \sum_{k=1}^{n} \frac{1}{\sqrt{k}}

2(\sqrt{n + 1} - 1) < 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{n}}

習題:

試以 M.I. 和非 M.I. 的手法,證明:

\sqrt[n]{a_1a_2 \dots a_n} + \sqrt[n]{b_1b_2 \dots b_n} \le \sqrt[n]{(a_1 + b_1)(a_2 + b_2) \dots (a_n + b_n)}

(where a_i, b_i (i = 1,2, \dots ,n) be non-negative numbers)

2 則迴響 »

  1. 單"純"地諗係好難諗到…

    迴響 由 FTL — 2010/09/28 @ 4:29 下午 | 回覆

  2. Hi,

    It is indeed useful to expose students to other non-routine proving methods. Thanks for sharing!

    Cheers,
    Wen Shih

    迴響 由 Wen Shih — 2010/09/29 @ 5:39 上午 | 回覆


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