已知数列 $u_{1}, u_{2}, u_{3}, \dots$ 的前 $n$ 项和 $S_{n}$ 由 $S_{n}=a(3^{n})+bn+c$ 给出，其中 $a, b$ 和 $c$ 是常数且 $n\ge 1$。

1. Calculate the sum of the first 1, 2, and 3 terms of the sequence.

$S_1 = u_1 = 5$
$S_2 = u_1 + u_2 = 5 + 9 = 14$
$S_3 = u_1 + u_2 + u_3 = 14 + 33 = 47$

2. Set up a system of linear equations using the given formula for $S_n$.

Substitute $n = 1, 2, 3$ into $S_n = a(3^n) + bn + c$:
$3a + b + c = 5 \quad \dots (1)$
$9a + 2b + c = 14 \quad \dots (2)$
$27a + 3b + c = 47 \quad \dots (3)$

3. Eliminate $c$ to form two equations in terms of $a$ and $b$.

Subtract (1) from (2):
$6a + b = 9 \quad \dots (4)$

Subtract (2) from (3):
$18a + b = 33 \quad \dots (5)$

4. Solve for $a, b$, and $c$.

Subtract (4) from (5):
$12a = 24 \implies a = 2$

Substitute $a = 2$ into (4):
$6(2) + b = 9 \implies 12 + b = 9 \implies b = -3$

Substitute $a = 2$ and $b = -3$ into (1):
$3(2) + (-3) + c = 5 \implies 3 + c = 5 \implies c = 2$

1. Concisely find constants $a$ and $b$ required from part (i).

Using $u_n = S_n - S_{n-1}$ for the first few terms:

Subtracting the first equation from the second gives:
$12a = 24 \implies a = 2$
Substituting $a = 2$ into $6a + b = 9$ gives:
$12 + b = 9 \implies b = -3$

2. Relate the $(n+1)$-th term to the sum formula.

By definition of the sum of a sequence:
$u_{n+1} = S_{n+1} - S_n$

3. Substitute the general formula and known constants to express $u_{n+1}$ in the required form.

$u_{n+1} = [a(3^{n+1}) + b(n+1) + c] - [a(3^n) + bn + c]$
$u_{n+1} = a(3^{n+1} - 3^n) + b(n + 1 - n) + (c - c)$
Factorising the exponent term $3^{n+1} - 3^n$:
$u_{n+1} = a(3 \cdot 3^n - 1 \cdot 3^n) + b$
$u_{n+1} = 2a(3^n) + b$
Substitute the constants $a = 2$ and $b = -3$ from part (i):
$u_{n+1} = 2(2)(3^n) - 3$
$u_{n+1} = 4(3^n) - 3$
Comparing this with $u_{n+1} = A(3^n) + B$, we have shown the relationship holds with $A = 4$ and $B = -3$.

1. Determine the constants $A$ and $B$ from part (ii) using the given terms of the sequence.

To use the result from part (ii), we first need to find the constants $A$ and $B$ for $u_{n+1} = A(3^n) + B$.
Given $u_2 = 9$ (when $n=1$) and $u_3 = 33$ (when $n=2$):
$u_2 = 3A + B = 9$
$u_3 = 9A + B = 33$
Subtracting the first equation from the second gives $6A = 24 \implies A = 4$.
Substituting $A = 4$ into $3A + B = 9 \implies 12 + B = 9 \implies B = -3$.
Thus, $u_{r+1} = 4(3^r) - 3$.

2. Set up the summation expression by splitting it into a geometric series and a constant series.

Substitute $u_{r+1}$ into the target summation:
$\sum_{r=2}^{n} u_{r+1} = \sum_{r=2}^{n} [4(3^r) - 3]$
This can be separated into two components:
$4 \sum_{r=2}^{n} 3^r - \sum_{r=2}^{n} 3$
The series $\sum_{r=2}^{n} 3^r$ is a finite geometric progression.

• First term ($r=2$): $a = 3^2 = 9$
• Common ratio: $R = 3$
• Number of terms: $N = n - 2 + 1 = n - 1$

3. Apply the sum formula for a geometric progression and combine the components.

Using the geometric progression sum formula $S_N = \frac{a(R^N - 1)}{R - 1}$:
$4 \sum_{r=2}^{n} 3^r = 4 \left( \frac{9(3^{n-1} - 1)}{3 - 1} \right) = \frac{36}{2}(3^{n-1} - 1) = 18(3^{n-1} - 1)$
For the constant sequence part, summing $3$ exactly $(n - 1)$ times yields:
$\sum_{r=2}^{n} 3 = 3(n - 1)$
Subtracting this from the geometric sum yields the final answer:
$\sum_{r=2}^{n} u_{r+1} = 18(3^{n-1} - 1) - 3(n - 1)$
(Note: The question explicitly states you need not simplify your answer.)