A series is given by $\displaystyle\sum_{r=1}^{n}2(4-3x)^{r}$ where $x$ is constant.

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

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 $x$ is a constant, the ratio between consecutive terms is a constant that is independent of $r$. Hence, the series is geometric.

2. State the condition for the sum to infinity of a geometric series to exist

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

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

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

1. Substitute $x=1$ into the general term and simplify the algebraic expression inside the summation.

Substitute $x = 1$ into $2(4-3x)^r + (r+1)(2r+5)$:

$2(4-3(1))^r + (2r^2 + 5r + 2r + 5)$
$= 2(1)^r + 2r^2 + 7r + 5$
$= 2r^2 + 7r + 7$

2. Split the sum and adjust the indices to prepare for formula application.

Distribute the summation operator:
$\sum_{r=0}^{n-1} (2r^2 + 7r + 7) = 2\sum_{r=0}^{n-1} r^2 + 7\sum_{r=0}^{n-1} r + \sum_{r=0}^{n-1} 7$

Recognising that for $r=0$, the terms $r^2$ and $r$ evaluate to $0$, we can adjust the starting index to $1$ for those sums:
$= 2\sum_{r=1}^{n-1} r^2 + 7\sum_{r=1}^{n-1} r + \sum_{r=0}^{n-1} 7$

3. Evaluate the sums using the given formula and standard arithmetic progression sum.

To use the given formula $\sum_{r=1}^{n} r^2 = \frac{n}{6}(n+1)(2n+1)$, replace $n$ with $n-1$:
$\sum_{r=1}^{n-1} r^2 = \frac{(n-1)}{6}(n)(2(n-1)+1) = \frac{n(n-1)(2n-1)}{6}$

Similarly, using the arithmetic series sum $\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$:
$\sum_{r=1}^{n-1} r = \frac{n(n-1)}{2}$

The sum of a constant $7$ repeated $n$ times (from $0$ to $n-1$) is $7n$.
Substituting these into our expression:
$= 2 \left[ \frac{n(n-1)(2n-1)}{6} \right] + 7 \left[ \frac{n(n-1)}{2} \right] + 7n$

4. Factorise $n$ and simplify the remaining algebraic fraction into a single quadratic polynomial.

Extract the common factor $n$:
$= n \left[ \frac{(n-1)(2n-1)}{3} + \frac{7(n-1)}{2} + 7 \right]$

Create a common denominator of $6$ inside the bracket:
$= n \left[ \frac{2(2n^2 - 3n + 1)}{6} + \frac{21(n - 1)}{6} + \frac{42}{6} \right]$
$= n \left[ \frac{4n^2 - 6n + 2 + 21n - 21 + 42}{6} \right]$
$= n \left[ \frac{4n^2 + 15n + 23}{6} \right]$

Rearranging to match the target form $n(an^2 + bn + c)$:
$= n \left( \frac{2}{3}n^2 + \frac{5}{2}n + \frac{23}{6} \right)$

(Alternatively, evaluating via index substitution $k=r+1$ avoids $(n-1)$ algebra and yields the same result directly).