和与积之研究(又续)

李煌定理

已知：$n\in \mathbb {Z} ,n\geq 2$

已知：$x^{n}-px^{n-1}=q$之解为 $x_{1},x_{2},x_{3}\dots x_{n}$

• 则壹定满足李煌关系：${\frac {1}{x_{1}^{n-1}}}+{\frac {1}{x_{2}^{n-1}}}+{\frac {1}{x_{3}^{n-1}}}+\dots +{\frac {1}{x_{n}^{n-1}}}={\frac {(1-n)p}{q}}$

• 则壹定满足李煌关系：${\frac {1}{(x_{1}x_{2})^{n-1}}}+{\frac {1}{(x_{1}x_{3})^{n-1}}}+\dots +{\frac {1}{(x_{n-1}x_{n})^{n-1}}}={\frac {p^{2}{\frac {(n-1)(n-2)}{2}}}{q^{2}}}$

• 则壹定满足李煌关系：${\frac {1}{(x_{1}x_{2}x_{3})^{n-1}}}+{\frac {1}{(x_{1}x_{2}x_{4})^{n-1}}}+\dots +{\frac {1}{(x_{n-2}x_{n-1}x_{n})^{n-1}}}={\frac {p^{3}{\frac {(1-n)(n-2)(n-3)}{6}}}{q^{3}}}$(未查到有人发现，应该是李煌首次发现，并由李煌证明)

• 则壹定满足李煌关系：${\frac {1}{(x_{1}x_{2}x_{3}x_{4})^{n-1}}}+{\frac {1}{(x_{1}x_{2}x_{3}x_{5})^{n-1}}}+\dots +{\frac {1}{(x_{n-3}x_{n-2}x_{n-1}x_{n})^{n-1}}}={\frac {p^{4}{\frac {(n-1)(n-2)(n-3)(n-4)}{24}}}{q^{4}}}$(未查到有人发现，应该是李煌首次发现，并由李煌证明)

• 则壹定满足李煌关系：${\frac {1}{(x_{1}x_{2}x_{3}x_{4}x_{5})^{n-1}}}+{\frac {1}{(x_{1}x_{2}x_{3}x_{4}x_{6})^{n-1}}}+\dots +{\frac {1}{(x_{n-4}x_{n-3}x_{n-2}x_{n-1}x_{n})^{n-1}}}={\frac {p^{5}{\frac {(1-n)(n-2)(n-3)(n-4)(n-5)}{120}}}{q^{5}}}$(未查到有人发现，应该是李煌首次发现，并由李煌证明)

更为一般之李煌关系通项如下($k\geq 1,k\leq n-1,k\in \mathbb {Z}$)

• 则壹定满足李煌关系：${\frac {1}{(x_{1}x_{2}\dots x_{k})^{n-1}}}+{\frac {1}{(x_{1}x_{2}\dots x_{k-1}x_{k+1})^{n-1}}}+\dots +{\frac {1}{(x_{n-k+1}x_{n-k+2}\dots x_{n-1}x_{n})^{n-1}}}={\frac {(-1)^{k}{\binom {n-1}{k}}p^{k}}{q^{k}}}$(未查到有人发现，应该是李煌首次发现，并由李煌证明)

推论

• 李煌-二项式系数定理

已知：$n\in \mathbb {Z} ,n\geq 2$

已知：$x^{n}-x^{n-1}+1=0$之解为 $x_{1},x_{2},x_{3}\dots x_{n}$

已知：$k\geq 1,k\leq n-1,k\in \mathbb {Z}$

则存在二项式系数李煌计算形式

${\frac {1}{(x_{1}x_{2}\dots x_{k})^{n-1}}}+{\frac {1}{(x_{1}x_{2}\dots x_{k-1}x_{k+1})^{n-1}}}+\dots +{\frac {1}{(x_{n-k+1}x_{n-k+2}\dots x_{n-1}x_{n})^{n-1}}}={\binom {n-1}{k}}$

推论

• 李煌-二项式系数定理

已知：$n\in \mathbb {Z} ,n\geq 1$

已知：$x^{n+1}-x^{n}+1=0$之解为 $x_{1},x_{2},x_{3}\dots x_{n},x_{n+1}$

已知：$k\geq 1,k\leq n,k\in \mathbb {Z}$

则存在二项式系数李煌计算形式

${\frac {1}{(x_{1}x_{2}\dots x_{k})^{n}}}+{\frac {1}{(x_{1}x_{2}\dots x_{k-1}x_{k+1})^{n}}}+\dots +{\frac {1}{(x_{n-k+2}x_{n-k+3}\dots x_{n}x_{n+1})^{n}}}={\binom {n}{k}}$

来源

• 《南昌理工学院学报》.李煌

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