已知一个级数为 $\sum_{r=1}^{n} 2(4-3x)^r$，其中 $x$ 为常数。

1. Show that the ratio of consecutive terms is a constant.

Let the $r$-th term be $T_r = 2(4-3x)^r$.
$\frac{T_{r+1}}{T_r} = \frac{2(4-3x)^{r+1}}{2(4-3x)^r} = 4-3x$
Since the ratio of consecutive terms is a constant ($4-3x$) that is independent of $r$, the series is a geometric series.

2. State the condition for the sum to infinity to exist.

For the sum to infinity of a geometric series to exist, the common ratio $R$ must satisfy $|R| < 1$.
$|4-3x| < 1$

3. Solve the inequality to find the range of $x$.

$-1 < 4 - 3x < 1$
Subtract $4$ from all parts:
$-5 < -3x < -3$
Divide by $-3$ and reverse the inequality signs:
$\frac{5}{3} > x > 1$
$1 < x < \frac{5}{3}$

1. Substitute $x = 1$ and expand the terms inside the summation.

Given $x = 1$, the term $2(4-3x)^r$ simplifies to $2(4-3)^r = 2(1)^r = 2$.
Expanding the second part:
$(r+1)(2r+5) = 2r^2 + 5r + 2r + 5 = 2r^2 + 7r + 5$
So the expression inside the sum is:
$2 + 2r^2 + 7r + 5 = 2r^2 + 7r + 7$

2. Split the sum into separate components and apply the given formula for $\sum r^2$.

By linearity, the sum can be split as:
$\sum_{r=0}^{n-1} (2r^2 + 7r + 7) = 2\sum_{r=0}^{n-1} r^2 + 7\sum_{r=0}^{n-1} r + \sum_{r=0}^{n-1} 7$
We are given $\sum_{r=1}^{n} r^2 = \frac{n}{6}(n+1)(2n+1)$. Since the $r=0$ term is $0$, we can replace $n$ with $n-1$ in the formula:
$\sum_{r=0}^{n-1} r^2 = \sum_{r=1}^{n-1} r^2 = \frac{(n-1)}{6}((n-1)+1)(2(n-1)+1) = \frac{n(n-1)(2n-1)}{6}$

3. Evaluate the remaining standard sums and combine all parts.

Using the sum of an arithmetic progression, $\sum_{r=0}^{n-1} r = \frac{n(n-1)}{2}$.
Since there are $n$ terms from $r=0$ to $n-1$, the constant sum is $\sum_{r=0}^{n-1} 7 = 7n$.
Substitute these back into the total sum:
$\text{Sum} = 2\left[\frac{n(n-1)(2n-1)}{6}\right] + 7\left[\frac{n(n-1)}{2}\right] + 7n$

4. Factorise $\frac{n}{6}$ and simplify to the required quadratic form.

Factorising $\frac{n}{6}$ from all terms:
$\text{Sum} = \frac{n}{6} \left[ 2(n-1)(2n-1) + 21(n-1) + 42 \right]$
$= \frac{n}{6} \left[ 2(2n^2 - 3n + 1) + 21n - 21 + 42 \right]$
$= \frac{n}{6} \left[ 4n^2 - 6n + 2 + 21n + 21 \right]$
$= \frac{n}{6} \left( 4n^2 + 15n + 23 \right)$
Expressing this in the requested form $n(an^2 + bn + c)$:
$= n\left(\frac{2}{3}n^2 + \frac{5}{2}n + \frac{23}{6}\right)$