If $a_k = \frac{1}{k(k+1)}$ for $k = 1, 2, 3,...n$, then $\begin{bmatrix} \displaystyle\sum_{k=1}^{n} a_k \end{bmatrix}^2$ is equal to :

Question

( A ) $\frac{n^6}{(n+1)^6}$
( B ) $\frac{n^2}{(n+1)^2}$
( C ) $\frac{n^4}{(n+1)^4}$
( D ) $\frac{n}{n+1}$