观察下列等式： $\sum_{i=1}^{n} i = \frac{1}{2} n^{2} + \frac{1}{2} n$， $\sum_{i=1}^{n} i^{2} = \frac{1}{3} n^{3} + \frac{1}{2} n^{2} + \frac{1}{6} n$， $\sum_{i=1}^{n} i^{3} = \frac{1}{4} n^{4} + \frac{1}{2} n^{3} + \frac{1}{4} n^{2}$， $\sum_{i=1}^{n} i^{4} = \frac{1}{5} n^{5} + \frac{1}{2} n^{4} + \frac{1}{3} n^{3} - \frac{1}{30} n^{2}$， $\sum_{i=1}^{n} i^{5} = \frac{1}{6} n^{6} + \frac{1}{2} n^{5} + \frac{5}{12} n^{4} - \frac{1}{12} n^{2}$， $\sum_{i=1}^{n} i^{6} = \frac{1}{7} n^{7} + \frac{1}{2} n^{6} + \frac{1}{2} n^{5} - \frac{1}{6} n^{3} + \frac{1}{42} n^{2}$， $\cdots \sum_{i=1}^{n} i^{k} = a_{k+1} n^{k+1} + a_{k} n^{k} + a_{k-1} n^{k-1} + a_{k-2} n^{k-2} + \cdots + a_{1} n + a_{0}$， 可以推测，当 $k \geqslant 2$（$k \in \mathbf{N}^{*}$）时，$a_{k+1} = \frac{1}{k+1}$，$a_{k} = \frac{1}{2}$，$a_{k-1} =$______，$a_{k-2} =$______.