The following is a proof by mathematical induction that the inequality $$\sum _ { i = 1 } ^ { 2 n + 1 } \frac { 1 } { n + i } = \frac { 1 } { n + 1 } + \frac { 1 } { n + 2 } + \cdots + \frac { 1 } { 3 n + 1 } > 1$$ holds for all natural numbers $n$.
$\langle$Proof$\rangle$ For a natural number $n$, let $a _ { n } = \frac { 1 } { n + 1 } + \frac { 1 } { n + 2 } + \cdots + \frac { 1 } { 3 n + 1 }$. It suffices to show that $a _ { n } > 1$.
(1) When $n = 1$, $a _ { 1 } = \frac { 1 } { 2 } + \frac { 1 } { 3 } + \frac { 1 } { 4 } > 1$.
(2) Assume that $a _ { k } > 1$ when $n = k$. $$\begin{aligned} & \text{When } n = k + 1, \\ & \begin{aligned} a _ { k + 1 } & = \frac { 1 } { k + 2 } + \frac { 1 } { k + 3 } + \cdots + \frac { 1 } { 3 k + 4 } \\ & = a _ { k } + \left( \frac { 1 } { 3 k + 2 } + \frac { 1 } { 3 k + 3 } + \frac { 1 } { 3 k + 4 } \right) \end{aligned} \end{aligned}$$ □
On the other hand, $( 3 k + 2 ) ( 3 k + 4 )$ (나) $( 3 k + 3 ) ^ { 2 }$, so $$\frac { 1 } { 3 k + 2 } + \frac { 1 } { 3 k + 4 } > \text{ (다) }$$ Since $a _ { k } > 1$, $$a _ { k + 1 } > a _ { k } + \left( \frac { 1 } { 3 k + 3 } + \text{ (다) } \right) \text{ (가) } > 1$$ Therefore, by (1) and (2), $a _ { n } > 1$ for all natural numbers $n$.
What are the correct expressions for (가), (나), and (다) in the above proof? [3 points]
| (가) | (나) | (다) |
| (1) $\frac { 1 } { k + 1 }$ | $<$ | $\frac { 2 } { 3 k + 3 }$ |
| (2) $\frac { 1 } { k + 1 }$ | $>$ | $\frac { 2 } { 3 k + 3 }$ |
| (3) $\frac { 1 } { k + 1 }$ | $<$ | $\frac { 2 } { 3 k + 4 }$ |
| (4) $\frac { 2 } { k + 1 }$ | $>$ | $\frac { 2 } { 3 k + 4 }$ |
| (5) $\frac { 2 } { k + 1 }$ | $<$ | $\frac { 2 } { 3 k + 3 }$ |