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 \(u_3\small.\)

$$ \begin{eqnarray} u_2 &=& -\!3u_1+4 \\[6pt] &=& -\!3(7)+4 \\[6pt] &=& -\!21+4 \\[6pt] &=& -\!17 \\[14pt] u_3 &=& -\!3u_2+4 \\[6pt] &=& -\!3(-17)+4 \\[6pt] &=& 51+4 \\[6pt] &=& 55 \end{eqnarray} $$

Video solution by Clelland Maths 

This example will require us to solve a pair of simultaneous equations.

For simplicity, let's use \(u_1\) (rather than \(u_0\)) as the first term.

$$ \begin{eqnarray} u_2 &=& mu_1+c \\[6pt] 20 &=& 12m+c \:\:\:① \\[14pt] u_3 &=& mu_2+c \\[6pt] 26 &=& 20m+c \:\:\:② \end{eqnarray} $$

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

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

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

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

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

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

Video solution by Clelland Maths 

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 \(u_4\small.\)

$$ \begin{eqnarray} u_3 &=& mu_2-3 \\[6pt] u_3 &=& 11m-3\:\:\:① \\[14pt] u_4 &=& mu_3-3 \\[6pt] 35 &=& mu_3-3\:\:\:② \\[6pt] \end{eqnarray} $$

Equations \(①\) and \(②\) now need to be solved simultaneously. However, the \(mu_3\) term is unlike anything that you encountered in Nat 5, so it makes sense to substitute the expression for \(u_3\) in equation \(①\) into equation \(②\):

$$ \begin{gather} 35=m(11m-3)-3\\[8pt] 35=11m^2-3m-3\\[8pt] 11m^2-3m-38=0\\[8pt] (m-2)(11m+19)=0\\[8pt] m-2=0\:\:\small\textsf{or}\normalsize\:\:11m+19=0\\[8pt] m=2\:\:\small\textsf{or}\normalsize\:\:m=-\small\frac{19}{11}\normalsize \end{gather} $$

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

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

Note: We like the 'ac method' (also known as 'split the middle') to factorise a non-unitary trinomial. But if you tried using the ac method to factorise \(11m^2-3m-38\small,\) and ended up staring blankly at the number \(-418\small,\) we hope you realised that 'factorising by inspection' is actually the way to go here. \(11\) is prime, and \(38\) is semiprime (i.e. a product of two primes) so it's a lot easier than it might seem.

Video solution by Clelland Maths 

\(a=\frac{2}{3}\) and \(b=4\small.\) 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 \(\large\frac{b}{1-a}\normalsize = \large\frac{4}{1\,-\,\Large{}^{2}\!/\!_{3}\normalsize}\normalsize = \large\frac{4}{\Large{}^{1}\!/\!_{3}\normalsize}\normalsize = 4\times 3=12.\)

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 in exams.

Video solution by Clelland Maths 

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

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

(b)  Note that the word 'limit' is not used in this part of the question. You are expected to know that the question is actually asking whether or not a limit exists.

A limit does exist, because this is a linear recurrence relation and \(-1\lt 0.965\lt 1\small.\) 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 \(15\,000\small.\)

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

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

Video solution by Clelland Maths 

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

$$ \begin{eqnarray} u_1 &=& ku_0-5 \\[6pt] &=& 20k-5\\[14pt] u_2 &=& ku_1-5 \\[6pt] &=& k(20k-5)-5 \\[6pt] &=& 20k^2-5k-5 \\[6pt] \end{eqnarray} $$

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

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

The critical values are \(k=-1\) and \(k=\frac{5}{4}\small.\) 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 \(-1\lt k\lt \frac{5}{4}\small.\)

Video solution by Clelland Maths 

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.

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

Similarly:

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

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

Video solution by Clelland Maths 

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

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

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

Video solution by Clelland Maths 

(a)  Using the recurrence relation:

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

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

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

Video solution by Clelland Maths 

This example needs us to solve a pair of simultaneous equations.

For simplicity, let's use \(u_1\) (rather than \(u_0\)) as the first term.

$$ \begin{eqnarray} u_2 &=& mu_1+c \\[6pt] 9 &=& 6m+c \:\:\:① \\[14pt] u_3 &=& mu_2+c \\[6pt] 11 &=& 9m+c \:\:\:② \end{eqnarray} $$

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

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

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

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

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

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

Video solution by Clelland Maths 

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

$$ \begin{gather} u_7=\small\frac{2}{3}\,\normalsize u_6+8\\[6pt] 20=\small\frac{2}{3}\,\normalsize u_6+8\\[6pt] \small\frac{2}{3}\,\normalsize u_6=12\\[6pt] 2u_6=36\\[6pt] u_6=18\\[18pt] u_6=\small\frac{2}{3}\,\normalsize u_5+8\\[6pt] 18=\small\frac{2}{3}\,\normalsize u_5+8\\[6pt] \small\frac{2}{3}\,\normalsize u_5=10\\[6pt] 2u_5=30\\[6pt] u_5=15 \end{gather} $$

(b)  \(\large\frac{b}{1-a}\normalsize = \large\frac{8}{1\,-\,\Large{}^{2}\!/\!_{3}\normalsize}\normalsize = \large\frac{8}{\Large{}^{1}\!/\!_{3}\normalsize}\normalsize = 8\times 3=24\)

Video solution by Clelland Maths 

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 \(u_k\small.\)

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

Video solution by Clelland Maths 

(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 \(-1\lt \large\frac{1}{5}\normalsize\lt 1\small.\)

     (ii)  \(\large\frac{12}{1\,-\,\Large{}^{1}\!/\!_{5}\normalsize}\normalsize = \frac{12}{\Large{}^{4}\!/\!_{5}\normalsize} = 12\times\!\large\frac{5}{4}\normalsize=15\)

Video solution by Clelland Maths 

(a)  Using \(a=m\small,\) \(b=4\) and a limit of \(10\):

$$ \begin{gather} \textsf{Limit}=\small\frac{b}{1-a} \\[6pt] 10=\small\frac{4}{1-m} \\[6pt] 10(1\!-\!m)=4 \\[6pt] 10-10m=4 \\[6pt] 10m=6 \\[6pt] m=\small\frac{6}{10} \\[6pt] m=\small\frac{3}{5} \\[6pt] \end{gather} $$

(b)  We now know that the recurrence relation is \(u_{n+1}=\large\frac{3}{5}\normalsize u_n + 4,\) so we use it to find \(u_0.\)

$$ \begin{gather} u_1=\small\frac{3}{5}\normalsize u_0+4 \\[6pt] 19=\small\frac{3}{5}\normalsize u_0+4 \\[6pt] 15=\small\frac{3}{5}\normalsize u_0 \\[6pt] 15\times5=3u_0 \\[6pt] 75=3u_0 \\[6pt] u_0=25 \\[6pt] \end{gather} $$