# Quod Erat Demonstrandum

## 2010/10/06

### 教學碎念

Filed under: NSS,Pure Mathematics — johnmayhk @ 3:22 下午

1.

（他們笑我出卷容易，因自己功力有限，我求他們見諒。）

Resolve $\frac{2x + 3}{x^2(x+3)^2}$ into partial fractions.

$\frac{1}{x(x+3)} \equiv \frac{1}{3}(\frac{1}{x} - \frac{1}{x+3})$
$\frac{d}{dx}\frac{1}{x(x+3)} \equiv \frac{1}{3}\frac{d}{dx}(\frac{1}{x} - \frac{1}{x+3})$
$\frac{2x + 3}{x^2(x+3)^2} \equiv \frac{1}{3}(\frac{1}{x^2} - \frac{1}{(x+3)^2})$

2.

For integer $n$, prove that $9^n + 10^n < 11^n$ iff $n \ge 5$.

$11^n - 9^n$
$= (10+1)^n - (10-1)^n$
$= 2(C_1^n10^{n-1} + C_3^n10^{n-3} + C_5^n10^{n-5} + \dots)$

$= 2(C_1^n10^{n-1} + C_3^n10^{n-3} + C_5^n10^{n-5} + \dots)$
$< 2(C_1^n10^{n-1} + C_3^n10^{n-1} + C_5^n10^{n-1} + \dots)$
$= 2\times 10^{n-1}(C_1^n + C_3^n + C_5^n + \dots)$
$= 2\times 10^{n-1}(2^{n-1}) = 2^n10^{n-1}$

For sufficiently large $n$, we have

$1^n + 2^n + 3^n + 4^n + 5^n + 6^n + 7^n + 8^n + 9^n + 10^n < 11^n$

（估計上式成立於整數 $n \ge 7$

$\sum_{k=1}^mk^n < (m+1)^n$ for sufficiently large $n$

For integer $n$, prove that $9^n + 10^n < 11^n$ iff $n \ge 5$.

$9^n + 10^n$
$< 10^n + 10^n$
$= 2(10^n)$

set

$2(10^n) < 11^n$
$\Longleftrightarrow \log2 + n < n\log11$
$\Longleftrightarrow n > \frac{\log2}{\log11 - 1} \approx 7$

$n > 7$$9^n + 10^ < 11^n$ 成立。

$1^n + 2^n + 3^n + 4^n + 5^n + 6^n + 7^n + 8^n + 9^n + 10^n < 11^n$

$\sum_{k=1}^mk^n < (m+1)^n$

3.

$\frac{n^2(n+1)^2}{4} = (\frac{n(n+1)}{2})^2 = (1 + 2 + 3 + \dots + n)^2$ 證畢。