Higher Maths
Sequences

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.

(c)  This part of the question is asking us to find the limit and see if it is greater than \(15\,000.\)

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 \(u_k\small.\)

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

(a)  Using the recurrence relation:

$$ \begin{eqnarray} u_2 &=& \small\frac15\normalsize u_1+12 \\[6pt] &=& \small\frac15\normalsize(20)+12 \\[6pt] &=& 4+12\\[6pt] &=& 16 \end{eqnarray} $$

(b) (i)  A limit exists because the recurrence relation is linear and \(-1\lt \frac{1}{5}\lt 1.\)

(b) (ii)  \(\large\frac{b}{1-a}\normalsize = \large\frac{12}{1-\large\frac{1}{5}\normalsize}\normalsize = \frac{12}{\large\frac{4}{5}\normalsize} = 12\!\times\!\frac{5}{4}=15\)