数列 $u_1, u_2, u_3, \dots$ 的前 $n$ 项和为 $S_n$，其中 $S_n = \sum_{r=1}^{n} u_r$。已知 $S_n = An^2(n + 1) - Bn$ 其中 $A$ 和 $B$ 是非零常数。

1. Use the relationship $u_n = S_n - S_{n-1}$ to set up the equation for $u_n$.

$u_n = S_n - S_{n-1}$
$u_n = [An^2(n + 1) - Bn] - [A(n - 1)^2(n - 1 + 1) - B(n - 1)]$
$u_n = [An^3 + An^2 - Bn] - [An(n - 1)^2 - B(n - 1)]$

2. Expand and simplify the expression to find $u_n$.

$u_n = An^3 + An^2 - Bn - [An(n^2 - 2n + 1) - Bn + B]$
$u_n = An^3 + An^2 - Bn - [An^3 - 2An^2 + An - Bn + B]$
$u_n = An^3 + An^2 - Bn - An^3 + 2An^2 - An + Bn - B$
$u_n = 3An^2 - An - B$

1. Form the first equation in $A$ and $B$ using the first term.

Using the given sum formula for $n=1$:
$S_1 = A(1)^2(1 + 1) - B(1) = 2A - B$

Since the first term $u_1$ is equivalent to the sum of the first term $S_1$:
$2A - B = 1 \quad \text{--- (1)}$

2. Form the second equation using the sum of the first two terms.

Using the sum formula for $n=2$:
$S_2 = A(2)^2(2 + 1) - B(2) = 12A - 2B$

Since $S_2$ represents the sum of the first two terms ($u_1 + u_2$):
$S_2 = 1 + 9 = 10$

Equating the two expressions for $S_2$:
$12A - 2B = 10$
Dividing by 2 gives:
$6A - B = 5 \quad \text{--- (2)}$

3. Solve the simultaneous equations to find $A$ and $B$.

Subtracting equation (1) from equation (2):
$(6A - B) - (2A - B) = 5 - 1$
$4A = 4 \implies A = 1$

Substituting $A = 1$ into equation (1):
$2(1) - B = 1$
$B = 1$

(Shown)

1. Express the term inside the summation in terms of the sequence $u$.

From parts (i) and (ii), the $n$th term of the sequence is $u_n = 3n^2 - n - 1$.
Substitute $n = r - 1$:
$u_{r-1} = 3(r-1)^2 - (r-1) - 1$
$u_{r-1} = 3(r-1)^2 - r + 1 - 1$
$u_{r-1} = 3(r-1)^2 - r$
Thus, the expression inside the summation is exactly $u_{r-1}$.

2. Rewrite the summation using $S_{n-1}$, evaluating the first term separately.

We need to evaluate $\sum_{r=1}^{n} u_{r-1}$.
For $r=1$, the term is $3(0)^2 - 1 = -1$.
For $r \ge 2$, the terms are $u_1, u_2, \dots, u_{n-1}$.
$\sum_{r=1}^{n} u_{r-1} = -1 + \sum_{r=2}^{n} u_{r-1}$
$= -1 + \sum_{k=1}^{n-1} u_k$
Since $S_{n-1} = \sum_{k=1}^{n-1} u_k$, this becomes:
$= -1 + S_{n-1}$

3. Substitute the formula for $S_{n-1}$ to find the final expression.

From part (ii), $S_n = n^2(n+1) - n$.
Substitute $n$ with $n-1$:
$S_{n-1} = (n-1)^2(n-1+1) - (n-1)$
$S_{n-1} = n(n-1)^2 - n + 1$
$S_{n-1} = n(n^2 - 2n + 1) - n + 1$
$S_{n-1} = n^3 - 2n^2 + n - n + 1 = n^3 - 2n^2 + 1$
Substituting this back into the summation:
$\sum_{r=1}^{n} \left[ 3(r - 1)^2 - r \right] = -1 + (n^3 - 2n^2 + 1)$
$= n^3 - 2n^2$
This is in the required form $an^3 - bn^2$.