一个数据处理中心每天处理数据（以适当的单位）。设 $u_n$（其中 $n$ 是整数，$n \ge 1$）表示自 $1^{\text{st}}$ September 起每天处理的数据量。已知 $u_{n+1} = (1 + p)u_n \text{ ,}$ 其中 $p$ 是一个正实常数。

1. Express the ratio of consecutive terms using the given recurrence relation

$\frac{u_{n+1}}{u_n} = 1 + p$

2. Conclude that the constant ratio defines a geometric progression

Since $p$ is a constant, $1 + p$ is also a constant. Because the ratio between any two consecutive terms in the sequence is a constant, $\{u_n\}$ is a geometric progression with a common ratio of $1+p$.

1. Equate the sum of the data processed in the first 3 days to the given total.

The amount of data processed in the first 3 days are $u_1$, $u_2$, and $u_3$.
$u_1 + u_2 + u_3 = \frac{127}{36}(2^m)$
Since $u_1 = 2^m$ and each subsequent term is multiplied by $(1+p)$:
$2^m + 2^m(1+p) + 2^m(1+p)^2 = \frac{127}{36}(2^m)$

2. Cancel $2^m$ and simplify to form a quadratic equation in $p$.

Divide both sides by $2^m$ (since $2^m \neq 0$):
$1 + (1+p) + (1 + 2p + p^2) = \frac{127}{36}$
$p^2 + 3p + 3 = \frac{127}{36}$
$p^2 + 3p - \frac{19}{36} = 0$
Multiply the entire equation by 36 to remove the denominator:
$36p^2 + 108p - 19 = 0$

3. Solve the quadratic equation for $p$ and reject the invalid root.

Factorising the quadratic equation:
$(6p - 1)(6p + 19) = 0$
$p = \frac{1}{6} \quad \text{or} \quad p = -\frac{19}{6}$
Since $p$ is given in the question as a positive real constant, we reject $p = -\frac{19}{6}$.
$p = \frac{1}{6}$

1. Relate the long-run total to the sum to infinity of the geometric series and check its common ratio.

The total amount of data handled in the long run corresponds to the sum to infinity of the geometric progression.
The common ratio of this progression is $r = 1 + p$.
Since $p$ is given as a positive real constant ($p > 0$), the common ratio $r = 1 + p > 1$.
(Note: Substituting $p = \frac{1}{6}$ from part (b) to show $r = \frac{7}{6} > 1$ is also perfectly valid.)

2. Conclude on the existence of the limit based on the condition for convergence.

For a geometric series to converge and have a finite limit (sum to infinity), the common ratio must satisfy $|r| < 1$.
Since $r > 1$, the sum to infinity diverges.
Therefore, there is no limit to the total amount of data the centre can handle in the long run.

1. Determine the explicit formula for $u_r$.

Given $p = \frac{1}{4}$, the common ratio of the geometric progression is $1 + \frac{1}{4} = \frac{5}{4}$.
The first term is $u_1 = 2^m$.
$u_r = u_1 (1 + p)^{r-1} = 2^m \left(\frac{5}{4}\right)^{r-1}$

2. Calculate $v_4$ in terms of powers of 2.

For $r = 4$, $v_4 = u_4$.
$v_4 = 2^m \left(\frac{5}{4}\right)^3$
$v_4 = 2^m \left(\frac{125}{64}\right) = 2^m \left(\frac{125}{2^6}\right)$
$v_4 = 125(2^{m-6})$

3. Formulate an expression for $v_r$ where $r \ge 4$ using the arithmetic progression formula.

The recurrence relation $v_r = v_{r-1} - 25$ for $r \ge 5$ indicates that from the 4th term onwards, $v_r$ is an arithmetic progression with first term $v_4$ and common difference $d = -25$.
The number of terms from the 4th to the $r$-th term is $(r - 4 + 1) = r - 3$.
The $r$-th term can be found by adding $(r-4)$ common differences to $v_4$:
$v_r = v_4 + (r - 4)(-25)$

4. Substitute $v_4$ and factorise to obtain the required form.

$v_r = 125(2^{m-6}) - 25(r - 4)$
Factorising $25$ from the expression:
$v_r = 25[5(2^{m-6}) - (r - 4)]$
$v_r = 25[5(2^{m-6}) - r + 4]$
Comparing with the given form $k[5(2^{m-6}) - r + 4]$, the statement is shown to be true with $k = 25$.

1. Find the sum of the data processed in the first 3 days using the geometric progression.

For $1 \le r \le 3$, $v_r = u_r$, which is a geometric progression with first term $a = 2^m$ and common ratio $1 + p = 1 + \frac{1}{4} = \frac{5}{4}$.
The sum of the first 3 terms is:
$S_3 = \frac{2^m \left( \left(\frac{5}{4}\right)^3 - 1 \right)}{\frac{5}{4} - 1}$
$S_3 = \frac{2^m \left( \frac{125}{64} - 1 \right)}{\frac{1}{4}} = 4 \left( \frac{61}{64} \right) 2^m = \frac{61}{16} 2^m$

2. Find the sum of the data processed from the 4th to the 15th day using the arithmetic progression.

From part (d), $v_r = 25\left[5(2^{m-6}) - r + 4\right]$ for $r \ge 4$ (having deduced $k=25$).
This forms an arithmetic progression with 12 terms (from $r=4$ to $15$).
The first term of this block is $v_4 = 25\left[5(2^{m-6})\right] = \frac{125}{64} 2^m$.
The last term is $v_{15} = 25\left[5(2^{m-6}) - 15 + 4\right] = \frac{125}{64} 2^m - 275$.
The sum of these 12 terms is:
$S_{4-15} = \frac{12}{2} (v_4 + v_{15})$
$S_{4-15} = 6 \left( \frac{125}{64} 2^m + \frac{125}{64} 2^m - 275 \right) = 6 \left( \frac{125}{32} 2^m - 275 \right) = \frac{375}{16} 2^m - 1650$

3. Calculate the total amount of data processed up to the 15th day.

The total data processed up to the 15th day is the sum of $S_3$ and $S_{4-15}$:
$\text{Total} = \frac{61}{16} 2^m + \frac{375}{16} 2^m - 1650$
$\text{Total} = \frac{436}{16} 2^m - 1650 = \frac{109}{4}(2^m) - 1650$

1. Determine the long-term behaviour of $v_r$ and explain its practical implication.

For $r \ge 5$, $v_r$ is an arithmetic sequence with a negative common difference ($-25$). As $r$ increases, the terms will eventually become negative.

Since $v_r$ represents the amount of data processed each day, it cannot be negative. Thus, the model becomes invalid in the long run, and it is not meaningful to compute the long-run total.