The following proves by mathematical induction that for all natural numbers $n$ $$\left( 1 ^ { 2 } + 1 \right) \cdot 1 ! + \left( 2 ^ { 2 } + 1 \right) \cdot 2 ! + \cdots + \left( n ^ { 2 } + 1 \right) \cdot n ! = n \cdot ( n + 1 ) !$$ holds.
(1) When $n = 1$, (left side) = 2, (right side) = 2, so the given equation holds.
(2) Assume it holds when $n = k$: $$\begin{aligned}
\left( 1 ^ { 2 } + 1 \right) \cdot 1 ! & + \left( 2 ^ { 2 } + 1 \right) \cdot 2 ! + \cdots \\
& + \left( k ^ { 2 } + 1 \right) \cdot k ! = k \cdot ( k + 1 ) !
\end{aligned}$$ Now show that it holds when $n = k + 1$. $$\begin{aligned}
\left( 1 ^ { 2 } + 1 \right) \cdot 1 ! + \left( 2 ^ { 2 } + 1 \right) \cdot 2 ! + \cdots + \left( ( k+1 ) ^ { 2 } + 1 \right) \cdot ( k + 1 ) ! \\
= k \cdot ( k + 1 ) ! + \left\{ ( k + 1 ) ^ { 2 } + 1 \right\} \cdot ( k + 1 ) ! \\
= ( \text{(a)} ) \\
= ( k + 1 ) \cdot \text{(b)} \cdot ( k + 1 ) ! \\
= \text{(c)}
\end{aligned}$$ Therefore, it also holds when $n = k + 1$. Thus, the given equation holds for all natural numbers $n$.
Which expressions are correct for (a), (b), and (c) in the above proof? [3 points]
| $\underline { ( \text { a } ) }$ | $\underline { ( \text { b } ) }$ | $\underline { ( \text { c } ) }$ |
| $( 1 ) k \cdot ( k + 1 ) !$ | $k ^ { 2 } + 2 k + 1$ | $( k + 1 ) !$ | |
| $( 2 ) k \cdot ( k + 1 ) !$ | $k ^ { 2 } + 3 k + 2$ | $( k + 2 ) !$ | |
| $( 3 ) k \cdot ( k + 1 ) !$ | $k ^ { 2 } + 3 k + 2$ | $( k + 1 ) !$ | |
| $( 4 ) ( k + 1 ) \cdot ( k + 1 ) !$ | $k ^ { 2 } + 3 k + 2$ | $( k + 2 ) !$ | |
| $( 5 ) ( k + 1 ) \cdot ( k + 1 ) !$ | $k ^ { 2 } + 2 k + 1$ | $( k + 1 ) !$ | |