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 $n$.
For a natural number $n$, let $a _ { n } = \frac { 1 } { n + 1 } + \frac { 1 } { n + 2 } + \cdots + \frac { 1 } { 3 n + 1 }$. We need 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 when $n = k$, $a _ { k } > 1$. $$\begin{aligned} & \text{When } n = k + 1 \\ & \begin{aligned} a _ { k + 1 } & = \frac { 1 } { k + 2 } + \frac { 1 } { k + 3 } + \cdots + \frac { 1 } { 3 k + 4 } \\ & = a _ { k } + \left( \frac { 1 } { 3 k + 2 } + \frac { 1 } { 3 k + 3 } + \frac { 1 } { 3 k + 4 } \right) \end{aligned} \end{aligned}$$ On the other hand, $( 3 k + 2 ) ( 3 k + 4 )$ $\square$ (b) $( 3 k + 3 ) ^ { 2 }$, so $$\frac { 1 } { 3 k + 2 } + \frac { 1 } { 3 k + 4 } > \text{(c)}$$ Since $a _ { k } > 1$, $$a _ { k + 1 } > a _ { k } + \left( \frac { 1 } { 3 k + 3 } + \text{(c)} \right) \text{(a)} > 1$$ Therefore, by (1) and (2), $a _ { n } > 1$ for all natural numbers $n$.
What are the correct values for (a), (b), and (c) in the above proof? [3 points]

(a)	(b)	(c)
(1) $\frac { 1 } { k + 1 }$	$>$	$\frac { 2 } { 3 k + 3 }$
(2) $\frac { 1 } { k + 1 }$	$<$	$\frac { 2 } { 3 k + 3 }$
(3) $\frac { 1 } { k + 1 }$	$<$	$\frac { 4 } { 3 k + 3 }$
(4) $\frac { 2 } { k + 1 }$	$>$	$\frac { 4 } { 3 k + 3 }$
(5) $\frac { 2 } { k + 1 }$	$<$	$\frac { 1 } { k + 1 }$

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) $\frac { 2 } { k + 1 }$	$<$	$\frac { 2 } { 3 k + 3 }$

The following proves by mathematical induction that for all natural numbers $n$, $$\sum _ { k = 1 } ^ { n } ( 5 k - 3 ) \left( \frac { 1 } { k } + \frac { 1 } { k + 1 } + \frac { 1 } { k + 2 } + \cdots + \frac { 1 } { n } \right) = \frac { n ( 5 n + 3 ) } { 4 }$$ holds.
(1) When $n = 1$, (left side) $= 2$, (right side) $= 2$, so the given equation holds.
(2) Assume that when $n = m$, the equation holds: $$\begin{aligned} & \sum _ { k = 1 } ^ { m } ( 5 k - 3 ) \left( \frac { 1 } { k } + \frac { 1 } { k + 1 } + \frac { 1 } { k + 2 } + \cdots + \frac { 1 } { m } \right) \\ = & \frac { m ( 5 m + 3 ) } { 4 } \end{aligned}$$ Now we show that it holds when $n = m + 1$. $$\begin{aligned} & \sum _ { k = 1 } ^ { m + 1 } ( 5 k - 3 ) \left( \frac { 1 } { k } + \frac { 1 } { k + 1 } + \cdots + \frac { 1 } { m + 1 } \right) \\ = & \sum _ { k = 1 } ^ { m } ( 5 k - 3 ) \left( \frac { 1 } { k } + \frac { 1 } { k + 1 } + \cdots + \frac { 1 } { m + 1 } \right) + \frac { \text { (가) } } { m + 1 } \\ = & \sum _ { k = 1 } ^ { m } ( 5 k - 3 ) \left( \frac { 1 } { k } + \frac { 1 } { k + 1 } + \cdots + \frac { 1 } { \sqrt { \text { (나) } } } \right) \\ \quad & \quad + \frac { 1 } { m + 1 } \sum _ { k = 1 } ^ { m } ( 5 k - 3 ) + \frac { \text { (가) } } { m + 1 } \\ = & \frac { m ( 5 m + 3 ) } { 4 } + \frac { 1 } { m + 1 } \sum _ { k = 1 } ^ { m + 1 } \text { (다) } \\ = & \frac { ( m + 1 ) ( 5 m + 8 ) } { 4 } \end{aligned}$$ Therefore, the equation also holds when $n = m + 1$. Thus, the given equation holds for all natural numbers $n$. What are the correct values for (가), (나), and (다) in the above proof? [4 points]

	(가)	(나)	(다)
(1)	$5 m - 3$	$m$	$5 k + 2$
(2)	$5 m - 3$	$m + 1$	$5 k + 2$
(3)	$5 m + 2$	$m$	$5 k - 3$
(4)	$5 m + 2$	$m$	$5 k + 2$
(5)	$5 m + 2$	$m + 1$	$5 k - 3$

The following is a proof by mathematical induction that for all natural numbers $n$,
$$\sum _ { k = 1 } ^ { n } ( 5 k - 3 ) \left( \frac { 1 } { k } + \frac { 1 } { k + 1 } + \frac { 1 } { k + 2 } + \cdots + \frac { 1 } { n } \right) = \frac { n ( 5 n + 3 ) } { 4 }$$
holds.

(1) When $n = 1$, (left side) $= 2$, (right side) $= 2$, so the given equation holds.
(2) Assume that the equation holds when $n = m$:
$$\begin{aligned} & \sum _ { k = 1 } ^ { m } ( 5 k - 3 ) \left( \frac { 1 } { k } + \frac { 1 } { k + 1 } + \frac { 1 } { k + 2 } + \cdots + \frac { 1 } { m } \right) \\ = & \frac { m ( 5 m + 3 ) } { 4 } \end{aligned}$$
Now we show that it holds when $n = m + 1$.
$$\begin{aligned} & \sum _ { k = 1 } ^ { m + 1 } ( 5 k - 3 ) \left( \frac { 1 } { k } + \frac { 1 } { k + 1 } + \cdots + \frac { 1 } { m + 1 } \right) \\ = & \sum _ { k = 1 } ^ { m } ( 5 k - 3 ) \left( \frac { 1 } { k } + \frac { 1 } { k + 1 } + \cdots + \frac { 1 } { m + 1 } \right) + \frac { \text { (가) } } { m + 1 } \\ = & \sum _ { k = 1 } ^ { m } ( 5 k - 3 ) \left( \frac { 1 } { k } + \frac { 1 } { k + 1 } + \cdots + \frac { 1 } { \sqrt { \text { (나) } } } \right) \\ \quad & \quad + \frac { 1 } { m + 1 } \sum _ { k = 1 } ^ { m } ( 5 k - 3 ) + \frac { \text { (가) } } { m + 1 } \\ = & \frac { m ( 5 m + 3 ) } { 4 } + \frac { 1 } { m + 1 } \sum _ { k = 1 } ^ { m + 1 } \text { (다) } \\ = & \frac { ( m + 1 ) ( 5 m + 8 ) } { 4 } \end{aligned}$$
Therefore, the equation also holds when $n = m + 1$. Thus, the given equation holds for all natural numbers $n$.
What should be filled in for (가), (나), and (다) in the above proof? [4 points]

	(가)	(나)	(다)
(1)	$5 m - 3$	$m$	$5 k + 2$
(2)	$5 m - 3$	$m + 1$	$5 k + 2$
(3)	$5 m + 2$	$m$	$5 k - 3$
(4)	$5 m + 2$	$m$	$5 k + 2$
(5)	$5 m + 2$	$m + 1$	$5 k - 3$

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) Assuming it holds when $n = k$,
$$\begin{aligned} \left( 1 ^ { 2 } + 1 \right) \cdot 1 ! & + \left( 2 ^ { 2 } + 1 \right) \cdot 2 ! + \cdots \\ & + \left( k ^ { 2 } + 1 \right) \cdot k ! = k \cdot ( k + 1 ) ! \end{aligned}$$
We show that it holds when $n = k + 1$.
$$\begin{aligned} \left( 1 ^ { 2 } + 1 \right) \cdot 1 ! + \left( 2 ^ { 2 } + 1 \right) \cdot 2 ! + \cdots + \left( k ^ { 2 } + 1 \right) \cdot k ! + \left\{ ( k + 1 ) ^ { 2 } + 1 \right\} \cdot ( k + 1 ) ! \\ = ( \text{ (A) } ) + \left\{ ( k + 1 ) ^ { 2 } + 1 \right\} \cdot ( k + 1 ) ! \\ = ( k + 1 ) \cdot \text{((B))} \cdot ( k + 1 ) ! \\ = \text{ ((C)) } \end{aligned}$$
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]

	$\underline { ( \text{ (A) } ) }$	$\underline { ( \text{ (B) } ) }$	$\underline { ( \text{ (C) } ) }$
$( 1 ) k \cdot ( k + 1 ) !$	$k ^ { 2 } + 2 k + 1$	$( k + 1 ) !$
$( 2 ) k \cdot ( k + 1 ) !$	$k ^ { 2 } + 3 k + 2$	$( k + 2 ) !$
$( 3 ) k \cdot ( k + 1 ) !$	$k ^ { 2 } + 3 k + 2$	$( k + 1 ) !$
$( 4 ) ( k + 1 ) \cdot ( k + 1 ) !$	$k ^ { 2 } + 3 k + 2$	$( k + 2 ) !$
$( 5 ) ( k + 1 ) \cdot ( k + 1 ) !$	$k ^ { 2 } + 2 k + 1$	$( k + 1 ) !$

The sequence $\left\{ a _ { n } \right\}$ satisfies
$$\left\{ \begin{array} { l } a _ { 1 } = \frac { 1 } { 2 } \\ ( n + 1 ) ( n + 2 ) a _ { n + 1 } = n ^ { 2 } a _ { n } \quad ( n = 1,2,3 , \cdots ) \end{array} \right.$$
The following is a proof by mathematical induction that for all natural numbers $n$,
$$\sum _ { k = 1 } ^ { n } a _ { k } = \sum _ { k = 1 }^{n} \frac { 1 } { k ^ { 2 } } - \frac { n } { n + 1 }$$
holds. $\langle$Proof$\rangle$
(1) When $n = 1$, (left side) $= \frac { 1 } { 2 }$, (right side) $= 1 - \frac { 1 } { 2 } = \frac { 1 } { 2 }$, so (*) holds.
(2) Assume (*) holds when $n = m$: $$\sum _ { k = 1 } ^ { m } a _ { k } = \sum _ { k = 1 } ^ { m } \frac { 1 } { k ^ { 2 } } - \frac { m } { m + 1 }$$
Now show that (*) holds when $n = m + 1$.
$$\begin{aligned} & \sum _ { k = 1 } ^ { m + 1 } a _ { k } = \sum _ { k = 1 } ^ { m } \frac { 1 } { k ^ { 2 } } - \frac { m } { m + 1 } + a _ { m + 1 } \\ = & \sum _ { k = 1 } ^ { m } \frac { 1 } { k ^ { 2 } } - \frac { m } { m + 1 } + \square \text { (a) } a _ { m } \\ = & \sum _ { k = 1 } ^ { m } \frac { 1 } { k ^ { 2 } } - \frac { m } { m + 1 } \\ & \quad + \frac { m ^ { 2 } } { ( m + 1 ) ( m + 2 ) } \cdot \frac { ( m - 1 ) ^ { 2 } } { m ( m + 1 ) } \cdot \cdots \cdot \frac { 1 ^ { 2 } } { 2 \cdot 3 } a _ { 1 } \end{aligned}$$
Therefore, (*) also holds when $n = m + 1$. Thus, (*) holds for all natural numbers $n$.
Which expressions are correct for (a), (b), and (c) in the above proof? [3 points]
(1) $\dfrac{\text{(a)}}{m} \quad \dfrac{\text{(b)}}{(m+1)(m+2)} \quad \dfrac{\text{(c)}}{\frac{1}{(m+1)^2(m+2)}} \quad \dfrac{1}{(m+1)(m+2)^2}$
(2) $\dfrac{m}{(m+1)(m+2)} \quad \dfrac{m}{(m+1)^2(m+2)} \quad \dfrac{1}{(m+1)(m+2)}$
(3) $\dfrac{m^2}{(m+1)(m+2)} \quad \dfrac{1}{(m+1)^2(m+2)} \quad \dfrac{1}{(m+1)(m+2)^2}$
(4) $\dfrac{m^2}{(m+1)(m+2)} \quad \dfrac{1}{(m+1)^2(m+2)} \quad \dfrac{1}{(m+1)(m+2)}$
(5) $\dfrac{m^2}{(m+1)(m+2)} \quad \dfrac{m}{(m+1)^2(m+2)} \quad \dfrac{1}{(m+1)(m+2)^2}$