已知级数： $\sum_{r=1}^{n} 2(4-3x)^{r}$ 其中 $x$ 为常数。

1. Show that the ratio between consecutive terms is constant to explain why it is a geometric series.

Let the $r$-th term be $u_r = 2(4-3x)^r$.
$\frac{u_{r+1}}{u_r} = \frac{2(4-3x)^{r+1}}{2(4-3x)^r} = 4-3x$
Since the ratio $\frac{u_{r+1}}{u_r}$ is independent of $r$, the series is a geometric series with common ratio $4-3x$.

2. State the condition for the convergence of a geometric series.

For the sum to infinity to exist, the common ratio $r$ must satisfy strictly between $-1$ and $1$ ($|r| < 1$):
$|4-3x| < 1$

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

$-1 < 4-3x < 1$
$-5 < -3x < -3$
$\frac{5}{3} > x > 1$
$1 < x < \frac{5}{3}$

1. Substitute $x=1$ into the summation and expand the general term.

$\sum_{r=0}^{n-1} \big[2(4-3(1))^{r} + (r+1)(2r+5)\big]$\n$= \sum_{r=0}^{n-1} \big[2(1)^{r} + 2r^2 + 5r + 2r + 5\big]$\n$= \sum_{r=0}^{n-1} \big(2r^2 + 7r + 7\big)$

2. Split the summation and apply the sum formulas, substituting $n-1$ for $n$.

$= 2\sum_{r=0}^{n-1} r^2 + 7\sum_{r=0}^{n-1} r + \sum_{r=0}^{n-1} 7$\nSince the $r=0$ terms evaluate to zero for the first two sums, we have:\n$= 2\sum_{r=1}^{n-1} r^2 + 7\sum_{r=1}^{n-1} r + \sum_{r=0}^{n-1} 7$\nUsing the given formula $\sum_{r=1}^{n} r^{2} = \frac{n}{6}(n+1)(2n+1)$ with $n$ replaced by $n-1$, and the arithmetic series sum $\sum_{r=1}^{n-1} r = \frac{n(n-1)}{2}$:\n$= 2 \left[ \dfrac{n-1}{6}(n-1+1)(2(n-1)+1) \right] + 7 \left[ \dfrac{n(n-1)}{2} \right] + 7n$\n$= 2 \left[ \dfrac{n(n-1)(2n-1)}{6} \right] + \dfrac{7n(n-1)}{2} + 7n$

3. Factor out $n$ and expand the terms inside the brackets.

$= n \left[ \dfrac{2(2n^2 - 3n + 1)}{6} + \dfrac{7(n-1)}{2} + 7 \right]$\n$= n \left[ \dfrac{2n^2 - 3n + 1}{3} + \dfrac{7n - 7}{2} + 7 \right]$

4. Group like terms to express the final answer in the required $n(an^2 + bn + c)$ form.

$= n \left( \dfrac{2}{3}n^2 - n + \dfrac{1}{3} + \dfrac{7}{2}n - \dfrac{7}{2} + 7 \right)$\n$= n \left[ \dfrac{2}{3}n^2 + \left(-1 + \dfrac{7}{2}\right)n + \left(\dfrac{1}{3} - \dfrac{7}{2} + 7\right) \right]$\n$= n \left( \dfrac{2}{3}n^2 + \dfrac{5}{2}n + \dfrac{23}{6} \right)$