# Quod Erat Demonstrandum

## 2015/01/09

### 推特老題

Filed under: HKALE,Pure Mathematics — johnmayhk @ 3:26 下午
Tags: ,

１１
１２１
１３３１
１４６４１

$\frac{x^k}{(x+1)(2x+1)(3x+1)\dots (kx+1)}$

$\frac{x^k}{(x+1)(2x+1)(3x+1)\dots (kx+1)}=1+\frac{x^k-(x+1)(2x+1)(3x+1)\dots (kx+1)}{(x+1)(2x+1)(3x+1)\dots (kx+1)}$

$\frac{x^k-(x+1)(2x+1)(3x+1)\dots (kx+1)}{(x+1)(2x+1)(3x+1)\dots (kx+1)}\equiv \frac{a_1}{x+1}+\frac{a_2}{2x+1}+\dots +\frac{a_k}{kx+1}$

$x^k-(x+1)(2x+1)(3x+1)\dots (kx+1)$

$\equiv a_1(2x+1)(3x+1)\dots (kx+1)+a_2(x+1)(3x+1)\dots (kx+1)+a_3(x+1)(2x+1)(4x+1)\dots (kx+1)+\dots +a_k(x+1)(2x+1)\dots ((k-1)x+1)$ ………….. (*)

$a_r(x+1)(2x+1)\dots (kx+1)$，這裡只有 $k-1$ 個括弧，因為當中沒有了 $(rx+1)$

$(-\frac{1}{r})^k=a_r(-\frac{1}{r}+1)(-\frac{2}{r}+1)(-\frac{3}{r}+1)\dots (-\frac{r-1}{r}+1)(-\frac{r+1}{r}+1)(-\frac{r+2}{r}+1)\dots (-\frac{k}{r}+1)$

$\frac{(-1)^k}{r}=a_r(r-1)(r-2)\dots(3)(2)(1)(-1)(-2)\dots(-(k-r))=a_r(r-1)!(-1)^{k-r}(k-r)!$

$a_r=\frac{(-1)^rC^k_r}{k!}$

$\frac{x^k}{(x+1)(2x+1)(3x+1)\dots (kx+1)}$

$=\frac{1}{k!}-\frac{C^k_1}{k!(x+1)}+\frac{C^k_2}{k!(2x+1)}-\frac{C^k_3}{k!(3x+1)}+\dots+\frac{(-1)^rC^k_r}{k!(kx+1)}$

$\frac{n^5}{(n+1)(2n+1)(3n+1)(4n+1)(5n+1)}=\frac{1}{5!}(1-\frac{5}{n+1}+\frac{10}{2n+1}-\frac{10}{3n+1}+\frac{5}{4n+1}-\frac{1}{5n+1})$