观察下列等式：
$$\begin{gathered} \underset{i=1}{\overset{n}{\sum }}i=\frac{1}{2}{n}^2+\frac{1}{2}n \\ \underset{i=1}{\overset{n}{\sum i^2}}=\frac{1}{3}{n}^2+\frac{1}{2}{n}^2+\frac{1}{6}n, \\ \underset{i=1}{\overset{n}{\sum i^3}}=\frac{1}{4}{n}^2+\frac{1}{2}{n}^2+\frac{1}{4}{n}^2, \end{gathered}$$

$\underset{i=1}{\overset{n}{\sum }}{i}^4=\frac{1}{5}{n}^4+\frac{1}{2}{n}^4+\frac{1}{3}{n}^3-\frac{1}{30}n,$

$\frac{2}{3}{a}_n+n-4,b_n={\left(-1\right)}^n\left(a_n-3n+21\right),$……………………………………
$\underset{i=1}{\overset{n}{\sum }}{i}^n=a_{k+1}{n}^{k+2}+a_k{n}^k+a_{k-1}{n}^{k-1}+a_{k-2}{n}^{k-2}+\dots +a_{1}n+a_0,$可以推测，当 $x\ge 2\left(k\in N^{*}\right)$时， $a_{k+1}=\frac{1}{k+1},a_k=\frac{1}{2},a_{k-1}=$， $a_{k-2}$=。