Substituting \(n=0\) into the recurrence relation:

$$ \begin{eqnarray} u_1 &=& ku_0+4 \\[6pt] 7 &=& k(-1)+4 \\[6pt] 7 &=& -k+4 \\[6pt] 3 &=& -k \\[6pt] k &=& -3 \end{eqnarray} $$

Now we know that the recurrence relation is \(u_{n+1}=-3u_n + 4,\) so we use it to find \(u_2\) and hence \(u_3.\)

$$ \begin{eqnarray} u_2 &=& -3u_1+4 \\[6pt] &=& -3(7)+4 \\[6pt] &=& -21+4 \\[6pt] &=& -17 \\[12pt] u_3 &=& -3u_2+4 \\[6pt] &=& -3(-17)+4 \\[6pt] &=& 51+4 \\[6pt] &=& 55 \end{eqnarray} $$

This example will require the solution of a pair of simultaneous equations. For simplicity, let's use \(u_1\) as the first term.

$$ \begin{eqnarray} u_2 &=& mu_1+c \\[6pt] 20 &=& 12m+c \:\:\:① \\[12pt] u_3 &=& mu_2+c \\[6pt] 26 &=& 20m+c \:\:\:② \end{eqnarray} $$

We can eliminate \(c\) if we subtract the two equations:

$$ \begin{eqnarray} 6 &=& 8m \\[6pt] m &=& \small\frac{6}{8}\normalsize \\[6pt] m &=& \small\frac{3}{4}\normalsize \end{eqnarray} $$

Now we just substitute \(m=\large\frac{3}{4}\normalsize\) back into either of the equations to find \(c.\) Using equation ①:

$$ \begin{eqnarray} 20 &=& 12m+c \\[6pt] 20 &=& 12\small\left(\frac{3}{4}\right)\normalsize+c \\[6pt] 20 &=& 9+c \\[6pt] c &=& 11 \\[6pt] \end{eqnarray} $$

We now know that the recurrence relation is \(u_{n+1}=\large\frac{3}{4}\normalsize u_n + 11,\) so we use it to find \(u_4.\)

$$ \begin{eqnarray} u_4 &=& \small\frac{3}{4}\normalsize u_3+11 \\[6pt] &=& \small\frac{3}{4}\normalsize \left(26\right)+11 \\[6pt] &=& \small\frac{39}{2}\normalsize+11 \\[6pt] &=& \small\frac{39}{2}\normalsize+\small\frac{22}{2}\normalsize \\[6pt] &=& \small\frac{61}{2}\normalsize \end{eqnarray} $$

This is a harder example requiring the solution of a pair of simultaneous equations. We will use \(u_3\) to bridge the gap between the given terms \(u_2\) and \(u_4.\)

$$ \begin{eqnarray} u_3 &=& mu_2-3 \\[6pt] u_3 &=& 11m-3\:\:\:① \\[12pt] u_4 &=& mu_3-3 \\[6pt] 35 &=& mu_3-3 \\[6pt] 38 &=& mu_3 \\[6pt] u_3 &=& \small\frac{38}{m}\normalsize\:\:\:② \\[6pt] \end{eqnarray} $$

Equations ① and ② contain two different ways of representing the intermediate term \(u_3\) so we can equate them and solve:

$$ \begin{eqnarray} 11m-3 &=& \small\frac{38}{m}\normalsize \\[6pt] 11m^2-3m &=& 38\\[6pt] 11m^2-3m-38 &=& 0\\[6pt] (m-2)(11m+19) &=& 0\\[6pt] \end{eqnarray} $$

$$ m-2=0\:\:\small\textsf{or}\normalsize\:\:11m+19=0 $$ $$ m=2\:\:\small\textsf{or}\normalsize\:\:m=-\small\frac{19}{11}\normalsize $$

We can disregard the fraction because we are told in the question that \(m\) is a positive integer. So \(m=2.\)

To find \(u_3,\) just substitute into either of the equations. For example, substituting into ②, \(u_3=\large\frac{38}{2}\normalsize = 19.\)

\(a=\small\frac{2}{3}\normalsize\) and \(b=4.\) The value of \(u_1\) is irrelevant for this question.

A limit exists because this is a linear recurrence relation and \(-1\lt a \lt 1.\)

The limit is \(\large\frac{b}{1-a}\normalsize = \large\frac{4}{1-\frac{2}{3}}\normalsize = \frac{4}{\frac{1}{3}} = 4\times 3=12.\)

Note that the formula for limit is not given in your formulae list. You will need to either learn it or know to solve \(L=\frac{2}{3} L+4.\) Either method is acceptable to the SQA.

(a)  A decline of \(3.5\%\) means that the population is being multiplied by \(100\%-3.5\%\) \(=96.5\%\) \(=0.965.\) So \(a=0.965.\)

\(b\) is the amount that the population is being increased by each year, so \(b=500.\)

(b)  The word "limit" is not used in this part of the question, but it is asking if a limit exists.

A limit does exist, because this is a linear recurrence relation and \(-1\lt 0.965\lt 1.\) So the population will stabilise in the long term.

Note that the \(u_0\) value of \(10\,000\) is irrelevant to this calculation.

The limit is \(\large\frac{b}{1-a}\normalsize = \large\frac{500}{1-0.965}\normalsize \approx 14\,286 .\) So the population will never exceed \(15\,000.\)

(a)  We use the recurrence relation twice to find and simplify an expression for \(u_2.\)

$$ \begin{eqnarray} u_1 &=& ku_0-5 \\[6pt] &=& 20k-5\\[12pt] u_2 &=& ku_1-5 \\[6pt] &=& k(20k-5)-5 \\[6pt] &=& 20k^2-5k-5 \\[6pt] \end{eqnarray} $$

(a)  Now we use the answer to (a) to set up a quadratic inequality:

$$ \begin{eqnarray} u_2 &\lt& u_0 \\[6pt] 20k^2-5k-5 &\lt& 20 \\[6pt] 20k^2-5k-25 &\lt& 0 \\[6pt] 5(4k^2-k-5) &\lt& 0 \\[6pt] 4k^2-k-5 &\lt& 0 \\[6pt] (k+1)(4k-5) &\lt& 0 \\[6pt] \end{eqnarray} $$

The critical values are \(k=-1\) and \(k=\frac{5}{4}.\) To solve the inequality we can either sketch the graph of \(y=(k+1)(4k-5)\) and look at when \(y\lt 0\) or we can construct a table of values, like this:

$$ \begin{array} {|c|c|} \hline k & \rightarrow & -1 & \rightarrow & \frac{5}{4} & \rightarrow \\ \hline (k+1)(4k-5) & + & 0 & - & 0 & + \\ \hline \end{array} $$

(Sketching the graph is quicker, by the way! It only needs to be a very rough sketch.)

So the range of values of \(k\) for which \(u_2\lt u_0\) is \(-1\lt k\lt \frac{5}{4}.\)

For convergence to a limit, \(-1\lt 3-k\lt 1\small.\)

We split this 'double less than' inequality into its two parts and solve each of them separately.

$$ \begin{eqnarray} -1 &\lt& 3-k \\[6pt] -1-3 &\lt& -\!k\\[6pt] -4 &\lt& -\!k \\[6pt] 4 &\gt& k \\[6pt] k &\lt& 4 \\[6pt] \end{eqnarray} $$

Similarly:

$$ \begin{eqnarray} 3-k &\lt& 1 \\[6pt] -k &\lt& 1-\!3 \\[6pt] -k &\lt& -\!2 \\[6pt] k &\gt& 2 \end{eqnarray} $$

Putting these two solutions back together, we can see that a limit only exists if \(2\lt k\lt 4\small.\)

(a)  \(u_{4}=\frac{1}{3} u_3+10 = \frac{1}{3}\left(6\right)+10 = 12\)

(a)  A limit exists because the recurrence relation is linear and \(-1\lt \frac{1}{3} \lt 1.\)

(c)  \(\large\frac{b}{1-a}\normalsize = \large\frac{10}{1-\frac{1}{3}}\normalsize = \frac{10}{\frac{2}{3}} = 10\!\times\!\frac{3}{2}=15\)

(a)  Using the recurrence relation:

$$ \begin{eqnarray} u_2 &=& mu_1+6 \\[6pt] 13 &=& 28m+6 \\[6pt] 13-6 &=& 28m\\[6pt] 28m &=& 7\\[6pt] m &=& \small\frac{7}{28}\normalsize \\[6pt] m &=& \small\frac{1}{4}\normalsize \end{eqnarray} $$

(b) (i)  A limit exists because the recurrence relation is linear and \(-1\lt \frac{1}{4}\lt 1.\)

(b) (ii)  \(\large\frac{b}{1-a}\normalsize = \large\frac{6}{1-\frac{1}{4}}\normalsize = \frac{6}{\frac{3}{4}} = 6\!\times\!\frac{4}{3}=8\)

This example will require the solution of a pair of simultaneous equations. For simplicity, let's use \(u_1\) as the first term.

$$ \begin{eqnarray} u_2 &=& mu_1+c \\[6pt] 9 &=& 6m+c \:\:\:① \\[12pt] u_3 &=& mu_2+c \\[6pt] 11 &=& 9m+c \:\:\:② \end{eqnarray} $$

We can eliminate \(c\) if we subtract the two equations:

$$ \begin{eqnarray} 2 &=& 3m \\[6pt] m &=& \small\frac{2}{3}\normalsize \end{eqnarray} $$

Now we just substitute \(m=\large\frac{2}{3}\normalsize\) back into either of the equations to find \(c.\) Using equation ①:

$$ \begin{eqnarray} 9 &=& 6m+c \\[6pt] 9 &=& 6\small\left(\frac{2}{3}\right)\normalsize+c \\[6pt] 9 &=& 4+c \\[6pt] c &=& 5 \\[6pt] \end{eqnarray} $$

(b)  We now know that the recurrence relation is \(u_{n+1}=\large\frac{2}{3}\normalsize u_n + 5,\) so we use it to find \(u_4.\)

$$ \begin{eqnarray} u_4 &=& \small\frac{2}{3}\normalsize u_3+5 \\[6pt] &=& \small\frac{2}{3}\normalsize \left(11\right)+5 \\[6pt] &=& \small\frac{22}{3}\normalsize+5 \\[6pt] &=& \small\frac{22}{3}\normalsize+\small\frac{15}{3}\normalsize \\[6pt] &=& \small\frac{37}{3}\normalsize \end{eqnarray} $$

(a)  We need to apply the recurrence relation twice to step backwards to \(u_5\small.\)

$$ \begin{eqnarray} u_7 &=& \small\frac{2}{3}\,\normalsize u_6+8 \\[6pt] 20 &=& \small\frac{2}{3}\,\normalsize u_6+8 \\[6pt] 20-8 &=& \small\frac{2}{3}\,\normalsize u_6 \\[6pt] 12 &=& \small\frac{2}{3}\,\normalsize u_6 \\[6pt] u_6 &=& 12\div\small\frac{2}{3}\normalsize \\[6pt] &=& 12\times\small\frac{3}{2}\normalsize \\[6pt] &=& 18 \\[14pt] u_6 &=& \small\frac{2}{3}\,\normalsize u_5+8 \\[6pt] 18 &=& \small\frac{2}{3}\,\normalsize u_5+8 \\[6pt] 18-8 &=& \small\frac{2}{3}\,\normalsize u_5 \\[6pt] 10 &=& \small\frac{2}{3}\,\normalsize u_5 \\[6pt] u_5 &=& 10\div\small\frac{2}{3}\normalsize \\[6pt] &=& 10\times\small\frac{3}{2}\normalsize \\[6pt] &=& 15 \end{eqnarray} $$

(b)  \(\large\frac{b}{1-a}\normalsize = \large\frac{8}{1-\frac{2}{3}}\normalsize = \frac{8}{\frac{1}{3}} = 8\!\times\!\frac{3}{1}=24\)

This question is a lot easier than it might first appear! We just set up a simple equation using the fact that the difference between the two terms is \(1000\) and then solve it to find \(u_k\small.\)

$$ \begin{gather} u_{k+1}-u_{k}=1000 \\[6pt] (9u_{k}-440)-u_{k}=1000 \\[6pt] 8u_{k}-440=1000 \\[6pt] 8u_{k}=1440 \\[6pt] u_{k}=\small\frac{1440}{8}\normalsize \\[6pt] u_{k}=180 \\[6pt] \end{gather} $$