Sum of Difference Squared for Sequence Terms

Given that the sum of the first $n$ terms of a sequence $\{a_k\}$ is given by: $S_n = \sum_{k=1}^n a_k = 6n^3$

The $k$-th term of the sequence, $a_k$, for $k \ge 1$ can be expressed in terms of the sum function as: $a_k = S_k - S_{k-1} \quad \text{for } k \ge 2, \text{ and } a_1 = S_1 = 6(1)^3 = 6$

Substituting the given formula for $S_k$: $a_k = 6k^3 - 6(k-1)^3$ $a_k = 6 \left[ k^3 - (k^3 - 3k^2 + 3k - 1) \right]$ $a_k = 6(3k^2 - 3k + 1)$ $a_k = 18k^2 - 18k + 6$

Now, we calculate the difference between consecutive terms, $a_{k+1} - a_k$: $a_{k+1} - a_k = \left[ 18(k+1)^2 - 18(k+1) + 6 \right] - \left[ 18k^2 - 18k + 6 \right]$ $a_{k+1} - a_k = 18 \left[ (k+1)^2 - k^2 \right] - 18 \left[ (k+1) - k \right]$ $a_{k+1} - a_k = 18(2k + 1) - 18(1)$ $a_{k+1} - a_k = 36k + 18 - 18 = 36k$

Dividing both sides by 36: $\frac{a_{k+1} - a_k}{36} = k$

We need to evaluate the required sum: $\sum_{k=1}^6 \left( \frac{a_{k+1} - a_k}{36} \right)^2 = \sum_{k=1}^6 k^2$

Using the formula for the sum of the squares of the first $N$ natural numbers, $\sum_{k=1}^N k^2 = \frac{N(N+1)(2N+1)}{6}$: $\sum_{k=1}^6 k^2 = \frac{6 \times (6 + 1) \times (2 \times 6 + 1)}{6} = \frac{6 \times 7 \times 13}{6} = 7 \times 13 = 91$