Given that $\sum_{r=1}^{n} r^{2} = \frac{n(n+1)(2n+1)}{6},$ find $\sum_{r=1}^{n} (3r+1)^{2}$ in terms of n. Use your answer to part (i) to find $\sum_{r=1}^{n+1} (3r-2)^{2}.$

1. Expand $(3r+1)^2$ and apply the linearity of the summation operator.

2. Substitute the standard summation formulas.

3. Factor out $\frac{n}{2}$ and simplify the resulting expression.

1. Compute the result of part (i) as it is required to solve part (ii).

$\sum_{r=1}^{n} (3r+1)^2 = \sum_{r=1}^{n} (9r^2 + 6r + 1)$
$= 9 \left( \frac{n(n+1)(2n+1)}{6} \right) + 6 \left( \frac{n(n+1)}{2} \right) + n$
$= \frac{3n(n+1)(2n+1)}{2} + 3n(n+1) + n$
$= \frac{n}{2} \left[ 3(2n^2 + 3n + 1) + 6(n+1) + 2 \right]$
$= \frac{n}{2} (6n^2 + 15n + 11)$

2. Relate the summation in part (ii) to the summation in part (i) by expanding the terms.

Expanding the series for part (ii):
$\sum_{r=1}^{n+1} (3r-2)^{2} = 1^2 + 4^2 + 7^2 + \dots + (3n+1)^2$
The sum of the terms from $4^2$ to $(3n+1)^2$ is exactly the series from part (i):
$\sum_{r=1}^{n} (3r+1)^2 = 4^2 + 7^2 + \dots + (3n+1)^2$
So we can write:
$\sum_{r=1}^{n+1} (3r-2)^{2} = 1^2 + \sum_{r=1}^{n} (3r+1)^2$

3. Substitute the result from part (i) into the relation.

$\sum_{r=1}^{n+1} (3r-2)^{2} = 1 + \frac{n}{2} (6n^2 + 15n + 11)$

4. Combine into a single fraction and factorise the polynomial completely.

$= \frac{2 + 6n^3 + 15n^2 + 11n}{2}$
$= \frac{6n^3 + 15n^2 + 11n + 2}{2}$
Testing $n = -1$ gives $6(-1)^3 + 15(-1)^2 + 11(-1) + 2 = -6 + 15 - 11 + 2 = 0$, so $(n+1)$ is a factor. Factorising the cubic polynomial yields:
$= \frac{(n+1)(6n^2 + 9n + 2)}{2}$