If $\displaystyle\sum_{k=1}^{n} a_k = 6n^3$, then $\displaystyle\sum_{k=1}^{6} \left(\dfrac{a_{k+1} - a_k}{36}\right)^2$ is equal to _______.

Given $\displaystyle\sum_{k=1}^{n} a_k = 6n^3$.

Let $S_n = 6n^3$. The $n$-th term of the sequence is given by $a_n = S_n - S_{n-1}$.

$a_n = 6n^3 - 6(n-1)^3$

$a_n = 6[n^3 - (n^3 - 3n^2 + 3n - 1)]$

$a_n = 6(3n^2 - 3n + 1)$

Replacing $n$ with $k$ and $k+1$, we get:

$a_k = 6(3k^2 - 3k + 1)$

$a_{k+1} = 6(3(k+1)^2 - 3(k+1) + 1) = 6(3k^2 + 3k + 1)$

Now, find the difference $a_{k+1} - a_k$:

$a_{k+1} - a_k = 6(3k^2 + 3k + 1) - 6(3k^2 - 3k + 1)$

$a_{k+1} - a_k = 6(6k) = 36k$

Substitute this into the given summation:

$\displaystyle\sum_{k=1}^{6} \left(\dfrac{a_{k+1} - a_k}{36}\right)^2 = \sum_{k=1}^{6} \left(\dfrac{36k}{36}\right)^2$

$\Rightarrow \displaystyle\sum_{k=1}^{6} k^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2$

Using the formula for the sum of squares of the first $n$ natural numbers, $\displaystyle\sum_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6}$:

$\displaystyle\sum_{k=1}^{6} k^2 = \dfrac{6(6+1)(2(6)+1)}{6} = \dfrac{6 \times 7 \times 13}{6} = 91$

Answer: $91$