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) $\dfrac{m^2}{(m+1)(m+2)} \quad \dfrac{m}{(m+1)^2(m+2)} \quad \dfrac{1}{(m+1)(m+2)^2}$