已知一个级数的前 $n$ 项和为 $S_n = 3n(n + 2)$。证明该级数的各项构成等差数列。 (a) 部分中该级数的第二项、第七项和第 $m$ 项是一个等比级数的前三个连续项，且其第 $n$ 项由 $v_n$ 给出。求 $m$ 的值，并解释该等比级数的无穷和是否存在。 另一个等比级数的第 $n$ 项由 $w_n = \mathrm{e}^{5 + nx(x + 1)}$ 给出，其中 $x$ 是常数。求 $x$ 的取值范围，使得该等比级数收敛。 在 (c) 部分中取 $x = -0.5$，且已知 (b) 部分中等比级数的前 $n$ 项和首次超过 (c) 部分中等比级数的第 $n$ 项，即 $\left(\sum_{r=1}^{n} v_r\right) > w_n$，求 $n$ 的最小值。

1. State the relationship between the $n$th term $u_n$ and the sum to $n$ terms $S_n$.

$u_n = S_n - S_{n-1}$

2. Substitute the expression for $S_n$ to find $u_n$ algebraically.

$u_n = 3n(n + 2) - 3(n - 1)(n - 1 + 2)$
$u_n = 3n^2 + 6n - 3(n - 1)(n + 1)$
$u_n = 3n^2 + 6n - 3(n^2 - 1)$
$u_n = 3n^2 + 6n - 3n^2 + 3$
$u_n = 6n + 3 \quad \text{for } n \ge 2$

Check for $n = 1$:
$u_1 = S_1 = 3(1)(1 + 2) = 9$
Since the formula $u_n = 6n + 3$ yields $6(1) + 3 = 9$, it is valid for all $n \ge 1$.

3. Show that the difference between consecutive terms is a constant to conclude it is an arithmetic progression.

$u_n - u_{n-1} = (6n + 3) - [6(n - 1) + 3]$
$u_n - u_{n-1} = 6n + 3 - (6n - 6 + 3)$
$u_n - u_{n-1} = 6$
Since $u_n - u_{n-1} = 6$, the difference between any two consecutive terms is a constant independent of $n$. Hence, the series follows an arithmetic progression.

1. Find the second, seventh, and $m$-th terms of the arithmetic series

From part (a), the first term of the AP is $a = S_1 = 3(1)(1 + 2) = 9$.
The sum of the first two terms is $S_2 = 3(2)(2 + 2) = 24$.
The second term is $u_2 = S_2 - S_1 = 24 - 9 = 15$.
The common difference is $d = u_2 - a = 15 - 9 = 6$.

The seventh term is:
$u_7 = a + 6d = 9 + 6(6) = 45$

The $m$-th term is:
$u_m = a + (m - 1)d = 9 + 6(m - 1) = 6m + 3$

2. Use the common ratio of the geometric series to find $m$

The first three consecutive terms of the geometric series are $v_1 = 15$, $v_2 = 45$, and $v_3 = 6m + 3$.

The common ratio $r$ is:
$r = \frac{v_2}{v_1} = \frac{45}{15} = 3$

Using the property of a geometric series, $v_3 = v_2 \times r$:
$6m + 3 = 45 \times 3$
$6m + 3 = 135$
$6m = 132$
$m = 22$

3. Determine if the sum to infinity exists by checking the condition for convergence

For the sum to infinity of a geometric series to exist, the common ratio $r$ must satisfy $|r| < 1$.
Since $r = 3$, the condition $|r| < 1$ is not met. Therefore, the sum to infinity does not exist.

1. Find the common ratio of the geometric series.

$w_n = \mathrm{e}^{5 + nx(x + 1)}$

The common ratio $r$ is given by:
$r = \frac{w_{n+1}}{w_n} = \frac{\mathrm{e}^{5 + (n+1)x(x + 1)}}{\mathrm{e}^{5 + nx(x + 1)}}$
$r = \mathrm{e}^{x(x + 1)}$

2. Set up the convergence inequality.

For a geometric series to converge, the common ratio must satisfy $-1 < r < 1$.

Since an exponential function is always positive, $\mathrm{e}^{x(x + 1)} > 0 > -1$, so we only need to satisfy $r < 1$.
$\mathrm{e}^{x(x + 1)} < 1$
$\mathrm{e}^{x(x + 1)} < \mathrm{e}^0$
$x(x + 1) < 0$

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

The roots of the equation $x(x + 1) = 0$ are $x = 0$ and $x = -1$.

Since the coefficient of $x^2$ is positive, it forms a U-shaped parabola. The expression is strictly negative between the roots.
$-1 < x < 0$

1. Determine the expressions for $\sum_{r=1}^{n} v_r$ and $w_n$ given $x = -0.5$.

From part (b), the geometric series $v_n$ has first term $v_1 = 15$ and common ratio $r = 3$.
The sum of the first $n$ terms is given by:
$\sum_{r=1}^{n} v_r = \frac{15(3^n - 1)}{3 - 1} = 7.5(3^n - 1)$
Substitute $x = -0.5$ into the expression for $w_n$ from part (c):
$w_n = \mathrm{e}^{5 + n(-0.5)(-0.5 + 1)} = \mathrm{e}^{5 - 0.25n}$

2. Set up the inequality and evaluate for integer values of $n$ to find the least $n$.

We require the sum to exceed $w_n$:
$7.5(3^n - 1) > \mathrm{e}^{5 - 0.25n}$
Since $n$ must be a positive integer, we can find the least value by checking small integer values sequentially or using a calculator table:

When $n = 2$:
LHS $= 7.5(3^2 - 1) = 60$
RHS $= \mathrm{e}^{5 - 0.25(2)} = \mathrm{e}^{4.5} \approx 90.017$
Since $60 < 90.017$, the inequality is not satisfied.

When $n = 3$:
LHS $= 7.5(3^3 - 1) = 195$
RHS $= \mathrm{e}^{5 - 0.25(3)} = \mathrm{e}^{4.25} \approx 70.105$
Since $195 > 70.105$, the inequality is satisfied.

The least value of $n$ is 3.