Eufisky - The lost book

## 一个与多项式分拆有关的级数题

\begin{align*}&x\left( {x - 1} \right)\left( {x - 2} \right) \cdots \left( {x - k + 1} \right) = {x^k} - {a_{k - 2}}x\left( {x - 1} \right)\left( {x - 2} \right) \cdots \left( {x - k + 2} \right) \\&-  \cdots  - {a_2}x\left( {x - 1} \right)\left( {x - 2} \right) - {a_1}x\left( {x - 1} \right) - {a_0}x.\end{align*}

$\frac{{{{\left( {1 + 2 + \cdots + \left( {k - 1} \right)} \right)}^2} - \left( {{1^2} + {2^2} + \cdots + {{\left( {k - 1} \right)}^2}} \right)}}{2} = \left( {1 + 2 + \cdots + \left( {k - 2} \right)} \right){a_{k - 2}} - {a_{k - 3}}.$故${a_{k - 3}} = \frac{{\left( {3k - 5} \right)\left( {k - 2} \right)\left( {k - 1} \right)k}}{{24}}.$

\begin{align*}&{\left( { - 1} \right)^{k - n}}\sum\limits_{1 \le {k_1} < {k_2} <  \cdots  < {k_{k - n}} \le k - 1} {{k_1}{k_2} \cdots {k_{k - n}}}  = {\left( { - 1} \right)^{k - n}}{a_{k - 2}}\sum\limits_{1 \le {k_1} < {k_2} <  \cdots  < {k_{k - n - 1}} \le k - 2} {{k_1}{k_2} \cdots {k_{k - n - 1}}} \\&+ {\left( { - 1} \right)^{k - n - 1}}{a_{k - 3}}\sum\limits_{1 \le {k_1} < {k_2} <  \cdots  < {k_{k - n - 2}} \le k - 3} {{k_1}{k_2} \cdots {k_{k - n - 2}}}  +  \cdots  + {\left( { - 1} \right)^2}{a_n}\sum\limits_{1 \le {k_1} \le n} {{k_1}} {x^n} + {\left( { - 1} \right)^1}{a_{n - 1}}.\end{align*}

\begin{align*}\sum\limits_{n = 1}^\infty  {\frac{{{n^k}}}{{n!}}}  &= \sum\limits_{n = 1}^\infty  {\frac{{{a_0}n + {a_1}n\left( {n - 1} \right) + {a_2}n\left( {n - 1} \right)\left( {n - 2} \right) +  \cdots  + {a_{k - 1}}n\left( {n - 1} \right)\left( {n - 2} \right) \cdots \left( {n - k + 1} \right)}}{{n!}}} \\&= {a_0}\sum\limits_{n = 1}^\infty  {\frac{1}{{\left( {n - 1} \right)!}}}  + {a_1}\sum\limits_{n = 2}^\infty  {\frac{1}{{\left( {n - 2} \right)!}}}  + {a_2}\sum\limits_{n = 3}^\infty  {\frac{1}{{\left( {n - 3} \right)!}}}  +  \cdots  + {a_{k - 1}}\sum\limits_{n = k}^\infty  {\frac{1}{{\left( {n - k} \right)!}}} \\&= \left( {{a_0} + {a_1} + {a_2} +  \cdots {a_{k - 1}}} \right)e = Ne.\end{align*}