Recurrence relations: DHS 2024 Promo Q12

Dunman High School2024 PromoPromo P1 Q12●●● Challenging12 marks

What this question tests

Recurrence Relations.

Question

. A nature conservationist group studies the trend of population growth of an endangered animal species over the years. Denoting $u_n$ as the population of the animals at the $n$ th year, the initial population in the first year is observed to be $u_1 = a$ , where $a \in \mathbb{Z}^+$ . The population at the $(n + 1)$ th year can be modelled by the relation $u_{n+1} = 2u_n - k, \qquad \text{where } k \text{ is a constant.}$

For parts (i) and (ii) below, it is given that $a = 4$ .
1. Describe what happens to the population over time if
  1. $k = 1$ , [1]
  2. $k = 5$ . [2]
2. Find the value of $k$ if there are $52$ animals in the $3$ rd year. [2]
Show that the sequence defined by $v_n = u_n - k$ is a geometric progression. By expressing $v_n$ in terms of $n$ or otherwise, show that $u_{n+1}$ can be written in the form $u_{n+1} = (a - k)2^n + k.$
State the value of $a$ such that the population will stay stabilized in the long term. [1]
The group intends to find out the sum of the population at every $5$ -year interval, starting from the $5$ th year and ending at the $80$ th year, i.e. $u_5 + u_{10} + \cdots + u_{80}$ . Give your answer in the form $A(2^B - 1)(a - k) + Ck$ , where $A$ , $B$ and $C$ are constants to be found. [3]

Show full worked solution▾

(a)(i)(A) $u_{n+1}=2u_n-k$ , $u_1=a=4$ ,

$k=1$ :

The population is always on the uptrend. Population will grow to infinity

(B) $u_1=4$ , $k=5$ $\begin{aligned} u_2&=2(4)-5=3\\ u_3&=2(3)-5=1\\ u_4&=2(1)-5=-3<0 \end{aligned}$ The population will become extinct by 4th year

(ii) $\begin{aligned} u_2&=2(4)-k=8-k\\ u_3&=2u_2-k=2(8-k)-k=16-3k=52\\ \therefore\ k&=-12 \end{aligned}$ (b) $v_n=u_n-k$ $\begin{aligned} v_{n+1}&=u_{n+1}-k\\ &=2u_n-2k=2(u_n-k)=2v_n \end{aligned}$ Thus $v_n$ is a geometric progression $\begin{aligned} v_n&=v_1(2)^{n-1}=(u_1-k)2^{n-1}=(a-k)2^{n-1}\\ u_n&=v_n+k=(a-k)2^{n-1}+k\\ u_{n+1}&=(a-k)2^n+k \end{aligned}$

Source marking note: amended parts

(c) $2^n\to\infty$ as $n\to\infty$

For population to stay stabilized as $n\to\infty$ , $a-k=0$ , i.e. $a=k$

(d) Rewriting $\begin{aligned} u_n&=(a-k)2^{n-1}+k\\ S&=(a-k)\left[2^4+2^9+\cdots+2^{79}\right]+16k\\ &=\frac{2^4\left[(2^5)^{16}-1\right]}{2^5-1}(a-k)+16k\\ &=\frac{16}{31}\left[2^{80}-1\right](a-k)+16k \end{aligned}$

The geometric-progression proof is written without division by vₙ, so it also covers the stabilised case a = k.

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