# Quod Erat Demonstrandum

## 2010/09/28

### 唔用 MI

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

Prove that

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

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

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。

“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。

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

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

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

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

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

（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 | 回應

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 | 回應