A company posted a video on a social media platform to advertise a new product. The video is uploaded at the start of $1^{\text{st}}$ April and the number of daily views is recorded at the end of each day. Let $u_n$, where $n \ge 1$, denotes the number of daily views recorded each day. It is given that $u_{n+1} = (1 + k)u_n,$ where $k$ is a positive constant.

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

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

2. Explain that a constant ratio defines a geometric progression.

Since $k$ is a constant, $1+k$ is also a constant. A sequence where the ratio of consecutive terms is a constant is a geometric progression.

1. Formulate an equation for the total number of views over the first 3 days using the sum of a geometric series.

The number of views forms a geometric progression with first term $a = u_1 = 311$ and common ratio $r = 1 + k$.
For the total views from $1^{\text{st}}$ April to $3^{\text{rd}}$ April, $n = 3$.
$S_3 = \frac{a(r^3 - 1)}{r - 1}$
$\frac{311((1+k)^3 - 1)}{(1+k) - 1} = 4043$

2. Expand and simplify the equation into a quadratic form.

$\frac{(1+k)^3 - 1}{k} = \frac{4043}{311}$
$\frac{1 + 3k + 3k^2 + k^3 - 1}{k} = 13$
$\frac{k(3 + 3k + k^2)}{k} = 13$
$k^2 + 3k + 3 = 13$
$k^2 + 3k - 10 = 0$

3. Solve the quadratic equation and determine the appropriate value for $k$.

$(k - 2)(k + 5) = 0$
$k = 2 \quad \text{or} \quad k = -5$
Since $k$ is given as a positive constant, we reject $k = -5$.
$k = 2$

1. Explain the divergence based on the condition for the sum to infinity of a geometric progression.

The common ratio of the geometric progression is $r = 1+k$.
Since $k$ is a positive constant (from the preamble) and $k=2$ (from part ii), we know that $r > 1$.
For a geometric progression to have a finite limit to its total sum in the long run (sum to infinity), the common ratio must satisfy the condition $|r| < 1$.
Because $r > 1$, this condition is not satisfied. Therefore, the series diverges and the total number of daily views increases without limit.

1. Calculate the total number of comments up to day 4 ($S_4$) and the number of comments on day 4 ($v_4$) using the geometric progression.

From part (ii), the common ratio of the geometric progression is $1+k = 3$, and $u_1 = 311$.
The number of comments on day 4 is:
$v_4 = u_4 = 311(3)^{4-1} = 8397$

The total number of comments up to day 4 is the sum of the first 4 terms of the GP:
$S_4 = \frac{311(3^4 - 1)}{3 - 1} = \frac{311(80)}{2} = 12440$

2. Determine the sum of the comments from day 5 to day $r$.

For $r \ge 5$, $v_r = v_{r-1} + 80$. This means the sequence of daily comments from day 5 onwards forms an arithmetic progression (AP) with a common difference $d = 80$.
The first term of this AP (on day 5) is:
$a = v_5 = v_4 + 80 = 8397 + 80 = 8477$

The number of terms from day 5 to day $r$ inclusive is $n = r - 4$.
The sum of comments from day 5 to day $r$ is:
$\text{Sum}_{\text{AP}} = \frac{n}{2} \left[ 2a + (n-1)d \right]$
$\text{Sum}_{\text{AP}} = \frac{r-4}{2} \left[ 2(8477) + (r - 4 - 1)(80) \right]$
$= \frac{r-4}{2} \left[ 16954 + 80(r - 5) \right]$
$= \frac{r-4}{2} \left[ 16954 + 80r - 400 \right]$
$= \frac{r-4}{2} \left[ 16554 + 80r \right]$
$= (r-4)(8277 + 40r)$
$= 8277r + 40r^2 - 33108 - 160r$
$= 40r^2 + 8117r - 33108$

3. Find the total number of comments up to the $r$-th day and determine the constant $t$.

The total number of comments up to the $r$-th day is the sum of comments up to day 4 and the comments from day 5 to $r$:
$\text{Total} = S_4 + \text{Sum}_{\text{AP}}$
$= 12440 + (40r^2 + 8117r - 33108)$
$= 40r^2 + 8117r - 20668$

Comparing this to the given expression $40r^2 + 8117r - t$:
$t = 20668$

1. Determine the constant $t$ by evaluating the sequences from earlier parts.

From part (ii), the views sequence $u_n$ is a geometric progression. Its sum for the first 3 days gives:
$311 + 311(1+k) + 311(1+k)^2 = 4043 \implies (1+k)^2 + (1+k) - 12 = 0$
Taking the positive root, $1+k = 3 \implies k=2$, so $u_n$ has a common ratio of $3$.
From part (iv), $v_4 = u_4 = 311(3^3) = 8397$. The sum of the first 4 days is $S_4 = \frac{311(3^4-1)}{3-1} = 12440$.
For $r \ge 5$, $v_r$ is an arithmetic progression with first term $v_5 = 8397 + 80 = 8477$ and common difference $d = 80$.
The total number of comments up to day $r$ is:
$S_r = 12440 + \frac{r-4}{2} \left[ 2(8477) + (r-4-1)80 \right]$
$= 12440 + (r-4)(8477 + 40r - 200)$
$= 40r^2 + 8117r - 33108 + 12440$
$= 40r^2 + 8117r - 20668$
Comparing with $40r^2 + 8117r - t$, we find $t = 20668$.

2. Set up the inequality for the total number of comments and find the critical value of $r$.

We require the total number of comments to be greater than 100 000:
$40r^2 + 8117r - 20668 > 100 000$
$40r^2 + 8117r - 120668 > 0$
Finding the roots of the corresponding quadratic equation $40r^2 + 8117r - 120668 = 0$ using the quadratic formula:
$r = \frac{-8117 \pm \sqrt{8117^2 - 4(40)(-120668)}}{2(40)}$
$r = \frac{-8117 \pm \sqrt{85192569}}{80}$
$r \approx 13.91 \quad \text{or} \quad r \approx -216.84$

3. Deduce the least number of days required.

Since $r$ represents the number of days, $r > 0$. From the quadratic inequality ($>0$, opening upwards), we must have $r > 13.91$.
Since $r$ must be an integer, the least number of days required is $r = 14$.