Racquel Sanderson

March 31, 2025 · 7:14 am

Deriving the Quadratic Equation formula

My year 10 students have been learning how to complete the square with the idea of then deriving the quadratic equation formula.

The general equation for a quadratic is $y=ax^2+bx+c$

Completing the square,

$\begin{equation*}ax^2+bx+c\end{equation}$

Factorise out the leading coefficient (i.e. $a$ )

$\begin{equation*}a(x^2+\frac{bx}{a}+\frac{c}{a})\end{equation}$

Half the second term (i.e $\frac{b}{a}$ ) and subtract the square of the second term.

$\begin{equation*}a((x+\frac{b}{2a})^2-(\frac{b}{2a})^2+\frac{c}{a})\end{equation}$

$\begin{equation*}a((x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{c}{a})\end{equation}$

Simplify

$\begin{equation*}a((x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{4ac}{4a^2})\end{equation}$

$\begin{equation*}a((x+\frac{b}{2a})^2+\frac{-b^2+4ac}{4a^2})\end{equation}$

$\begin{equation*}a(x+\frac{b}{2a})^2+\frac{-b^2+4ac}{4a}\end{equation}$

Now let’s solve

$\begin{equation*}a(x+\frac{b}{2a})^2+\frac{-b^2+4ac}{4a}=0\end{equation}$

$\begin{equation*}a(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a}\end{equation}$

$\begin{equation*}(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}\end{equation}$

$\begin{equation*}(x+\frac{b}{2a})=\pm \sqrt{\frac{b^2-4ac}{4a^2}}\end{equation}$

$\begin{equation*}(x+\frac{b}{2a})=\frac{\pm \sqrt{b^2-4ac}}{2a}\end{equation}$

$\begin{equation*}x=-\frac{b}{2a}\frac{\pm \sqrt{b^2-4ac}}{2a}\end{equation}$

Which is the quadratic equation formula.

Leave a Comment

Filed under Algebra, Quadratic, Quadratics, Solving, Solving, Solving Equations

Tagged as completing the square, quadratic equation formula

March 26, 2025 · 12:23 pm

Logistic Growth Worked Example

A brumby is a free-roaming wild horse found in large number in parts of Australia. The culling of brumbies was banned in the year 2000. At this time the estimated population of brumbies in
Kosciuszko National Park was 1600. Scientists have modelled the population, P(t), of brumbies in
Kosciuszko National Park t years since the ban, by

$\begin{equation*}P(t)=\frac{18000}{10.25e^{0.15t}+1}\end{equation}$

(a) Use the model to determine how long it will take the brumbies to increase to a number that is triple the number when the ban came into effect.

(b) From this model, determine the estimated long run number of brumbies in Kosciuszko National Park.

It can be shown that the growth rate of the population of brumbies can be expressed as

$\begin{equation*}\frac{dP}{dt}=\frac{1}{r}P(k-P)\end{equation}$

(d) Determine the greatest growth rate for the population of brumbies.

ATAR 2024 Specialist Mathematics Question 13

(a) $1600\times 3=4800$
$4800=\frac{18000}{10.25e^{0.15t}+1}$
$t=8.8$ years.

(b) $\lim_{\limits_{t \to \infty}\frac{18000}{10.25e^{0.15t}+1}=18000$

$\begin{equation*}\frac{dP}{dt}=rP(k-P)\Longleftrightarrow P=\frac{KP_0}{(k-P_0)e^{-rkt}+P_0}\end{equation}$

We have $\frac{1}{r}$ instead of $r$ .

Therefore,

$\begin{equation*}0.15=\frac{1}{r}\times 18000\end{equation}$

$r=120000$

(d) The greatest growth rate occurs when $P=\frac{k}{2}=9000$

$\begin{equation*}\frac{dP}{dt}=\frac{1}{r}P(k-P)\end{equation}$

$\begin{equation*}\frac{dP}{dt}=\frac{1}{120000}(9000)(9000)\end{equation}$

The greatest growth rate is 675 Brumbies per year.

1 Comment

Filed under Differential Equations, Logistic Growth, Year 12 Specialist Mathematics

Tagged as differential equations, logisitic growth, worked example, Year 12 Specialist Mathematics

March 25, 2025 · 7:32 am

Deriving the Logistic Growth Equation

The logistic differential equation

$\begin{equation*}\frac{dP}{dt}=rP(k-P)\end{equation}$

where $r$ is the growth parameter and $k$ is the carrying capacity.

And the maximum rate of increase happens when $P=\frac{k}{2}$

$\begin{equation*}\frac{dP}{dt}=rP(k-P)\end{equation}$

$\begin{equation*}\frac{dP}{P(k-P)}=r dt{\end{equation}$

$\begin{equation*}\int \frac{dP}{P(k-P)}=\int r dt{\end{equation}$

I am going to separate the denominator on the left hand side

\frac{1}{P(k-P)}=\frac{A}{P}+\frac{B}{k-P}

\frac{1}{P(k-P)}=\frac{A(k-P)+BP}{P(k-P)}

So our equation is,

$\begin{equation*}\int \frac{\frac{1}{k}}{P}+\frac{\frac{1}{k}}{k-P} dP=\int r dt\end{equation}$

$\begin{equation*}\frac{1}{k}\int \frac{1}{P}+\frac{1}{k-P} dP=\int r dt\end{equation}$

$\begin{equation*}\int \frac{1}{P}+\frac{1}{k-P} dP=\int kr dt\end{equation}$

$\begin{equation*}ln\lvert{P}\rvert-ln\lvert{k-P}\rvert=krt+c\end{equation}$

$\begin{equation*}ln\lvert{\frac{P}{k-P}\rvert=krt+c\end{equation}$

$\begin{equation*}\frac{P}{k-P}=e^{krt+c}\end{equation}$

$\begin{equation*}\frac{P}{k-P}=e^{krt}e^{c} \end{equation}$

When $t=0, P=P_0$ ,

$\begin{equation*}\frac{P_0}{k-P_0}=e^{c} \end{equation}$

The equation is now

$\begin{equation*}\frac{P}{k-P}=\frac{P_0}{k-P_0}e^{krt}\end{equation}$

$\begin{equation*}P=\frac{P_0}{k-P_0}e^{krt}(k-P)\end{equation}$

$\begin{equation*}P=k\frac{P_0}{k-P_0}e^{krt}-P\frac{P_0}{k-P_0}e^{krt}\end{equation}$

$\begin{equation*}P+P\frac{P_0}{k-P_0}e^{krt}=k\frac{P_0}{k-P_0}e^{krt}\end{equation}$

$\begin{equation*}P(1+\frac{P_0}{k-P_0}e^{krt})=k\frac{P_0}{k-P_0}e^{krt}\end{equation}$

$\begin{equation*}P=\frac{k\frac{P_0}{k-P_0}e^{krt}}{1+\frac{P_0}{k-P_0}e^{krt}}\end{equation}$

$\begin{equation*}P=\frac{k\frac{P_0}{k-P_0}e^{krt}}{\frac{k-P_0+P_0e^{krt}}{k-P_0}}\end{equation}$

$\begin{equation*}P=\frac{kP_0e^{rkt}}{k-P_0+P_0e^{rkt}}\end{equation}$

Divide by $e^{rkt}$

$\begin{equation*}P=\frac{kP_0}{(k-P_0)e^{-rkt}+P_0}\end{equation}$

$\begin{equation*}}\frac{dP}{dt}=rP(k-P)\Longleftrightarrow P=\frac{kP_0}{(k-P_0)e^{-rkt}+P_0}\end{equation}$

Proving the Maximum Rate of Increase Happens When $P=\frac{k}{2}$

$\begin{equation*}\frac{dP}{dt}=rP(k-P)\end{equation}$

$\begin{equation*}\frac{d^2P}{dt^2}=r\frac{dP}{dt}(k-P)+rP(-\frac{dP}{dt})\end{equation}$

$\begin{equation*}\frac{d^2P}{dt^2}=\frac{dP}{dt}(rk-rP-rP)\end{equation}$

$\begin{equation*}\frac{d^2P}{dt^2}=0\end{equation}$

$\begin{equation*}\frac{dP}{dt}(rk-rP-rP)=0\end{equation}$

$\begin{equation*}r\frac{dP}{dt}(k-2P)=0\end{equation}$

$\begin{equation*}\frac{dP}{dt}(k-2P)=0\end{equation}$

$\begin{equation*}rP(k-P)(k-2P)=0\end{equation}$

Hence $P=k$ or $P=\frac{k}{2}$

(1) $\begin{equation*}\frac{d^3P}{dt^3}=\frac{dP^2}{dt^2}(rk-2rP)+\frac{dP}{dt}(-2\frac{dP}{dt})\end{equation*}$

Substitute $P=k$ into equation $1$

$\begin{equation*}\frac{d^3P}{dt^3}=rk(k-k)(rk-2rk)(rk-2rk)-2(rk(k-k))^2=0\end{equation}$

Hence, not a maximum.

Substitute $P=\frac{k}{2}$ into equation $1$

$\begin{equation*}\frac{d^3P}{dt^3}=rk(k-\frac{k}{2})(rk-2r\frac{k}{2})(rk-2r\frac{k}{2})-2(rk(k-\frac{k}{2}))^2=0\end{equation}$

$\begin{equation*}\frac{d^3P}{dt^3}=-2(rk^2-\frac{rk^2}{2})^2\end{equation}$

$\begin{equation*}\frac{d^3P}{dt^3}=-2\frac{r^2k^4}{4}\end{equation}$

$-2\frac{r^2k^4}{4}\le 0$ For all values of $P, r$ and $k$ .

Hence maximum when $P=\frac{k}{2}$

We will look at a worked example in the next post.

Leave a Comment

Filed under Differential Equations, Differentiation, Implicit, Logistic Growth, Optimisation, Product Rule, Uncategorized, Year 12 Specialist Mathematics

Tagged as differential equations, logistic growth, Optimisation, partial fractions

March 18, 2025 · 6:04 am

Intersecting Secant Theorem

$CD$ is a tangent to the circle.

Prove $c^2=a(a+b)$

I am going to add two chords to the circle

$\angle{BDC}=\angle{CAD}$ (angles in alternate segments are congruent)

$\angle{BCD}=\angle{DCA}$ (shared angle)

$\therefore \Delta BDC\cong \Delta{DAC}$ (AA)

Hence

$\frac{DC}{AC}=\frac{BC}{DC}$ (Corresponding sides in similar triangles)

$\frac{c}{a+b}=\frac{a}{c}$

$\therefore c^2=a(a+b)$

1 Comment

Filed under Circle Theorems, Geometry, Year 11 Specialist Mathematics

Tagged as circle geometry, Intersecting Secants Theorem

March 17, 2025 · 12:33 pm

Circle Geometry Question 2

One of my Year 11 Specialist students had this question

Triangle $ABC$ touches the given circle at Points $P, Q, R$ and $S$ only. The secant $BW$ touches the circle at $V$ and $W$ .

Diagram not drawn to scale

(a) Determine the lengths of the line segments marked $x, y$ and $z$ , leaving your answers as exact values.

(b) If the length of the line segment $QW$ is 4 units, determine the exact radius of the circle.

(a) We are going to use the Intersecting Secant Theorem – the tangent version

Hence, we have

$\begin{equation*}30^2=25(25+x+6)\end{equation}$

$\begin{equation*}900=25(31+x)\end{equation}$

$\begin{equation*}x=5\end{equation}$

Then we can use the intersecting chord theorem to find $y$ .

$\begin{equation*}10\times y=6 \times x\end{equation}$

$\begin{equation*}10y=30\end{equation}$

$\begin{equation*}y=3\end{equation}$

Back to the Intersecting Secant Theorem to find $z$

$\begin{equation*}z^2=4\times 17\end{equation}$

$\begin{equation*}z=2\sqrt{17}\end{equation}$

(b)

$QW$ is part of a 3-4-5 triangle, therefore $\angle{Q}=90^\circ$

This is definitely the case of the diagram not being drawn to scale. If $\angle{Q}=90^\circ$ , then the purple line must be the diameter.

We can use pythagoras to find the length of the diameter

$\begin{equation*}(2r)^2=13^2+4^2\end{equation}$

$\begin{equation*}4r^2=185\end{equation}$

$\begin{equation*}r=\frac{\sqrt{285}}{2}\end{equation}$

The radius of the circle is $\frac{\sqrt{285}}{2}$

Leave a Comment

Filed under Circle Theorems, Geometry, Pythagoras, Year 11 Specialist Mathematics

Tagged as intersecting chord theorem, intersecting secant theorem, pythagoras

March 9, 2025 · 11:26 am

Circle Geometry Question

In the above diagram $O$ is the centre of the larger circle. $A, B,D$ and $E$ are points on the circumference of the larger circle. $A, C, E$ and $0$ are points on the circumference of the smaller circle. Show that $\angle{CAB}=\angle{ABC}$ . $AB, AC$ and $BC$ are straight lines.

$AO=OB$ (radii of the larger circle)

At a line from $O$ to $E$ (it is also a radius of the larger circle)

Let $\angle{CAB}=\alpha$ .

$ACEO$ is a cyclic quadrilateral.

Hence, $\angle{CED}=180-\alpha$ ( $AECO$ is a cyclic quadrilateral)

As $CB$ is a straight line $\angle{OEB}=180-(180-\alpha)=\alpha$ .

$\Delta OEB$ is an isosceles triangle.

Therefore, $\angle{ABC}=\alpha$

Therefore $\angle{ABC}=\angle{CAB}$

Leave a Comment

Filed under Circle Theorems, Finding an angle, Geometry, Year 11 Specialist Mathematics

March 5, 2025 · 2:00 am

Binomial Expansion Theorem

My Year 11 Mathematics Methods students are working on the Binomial Expansion Theorem.

But before we get onto that, remember Pascal’s triangle

Now we can use combinations to find the numbers in each row. For example, $1 4 6 4 1$ is $\begin{pmatrix}4\\0\end{pmatrix}=1, \begin{pmatrix}4\\1\end{pmatrix}=4, \begin{pmatrix}4\\2\end{pmatrix}=6, \begin{pmatrix}4\\3\end{pmatrix}=4, \begin{pmatrix}4\\4\end{pmatrix}=1$

As you can see, the coefficients are the row of pascal’s triangle corresponding to the power. So $(x+y)^6$ would have co-efficients from the sixth row of the table $1, 6, 15, 20, 15, 6, 1$ .

To generalise

$(x+y)^n=\begin{pmatrix}n\\0\end{pmatrix}x^ny^0+\begin{pmatrix}n\\1\end{pmatrix}x^{n-1}y^1+\begin{pmatrix}n\\2\end{pmatrix}x^{n-2}y^2+ ...+\begin{pmatrix}n\\n-1\end{pmatrix}x^1{y^{n-1}+\begin{pmatrix}n\\n\end{pmatrix}x^0y^n$

Which we can condense to

$(x+y)^n=\Sigma_{i=0}^n \begin{pmatrix}n\\i\end{pmatrix}x^{n-i}y^i$

Worked Examples

$(1)$ Expand $(2x-3)^4$

$(2x-3)^4=\begin{pmatrix}4\\0\end{pmatrix}(2x)^4(-3)^0+\begin{pmatrix}4\\1\end{pmatrix}(2x)^3(-3)^1+\begin{pmatrix}4\\2\end{pmatrix}(2x)^2(-3)^2+\begin{pmatrix}4\\3\end{pmatrix}(2x)^1(-3)^3+\begin{pmatrix}4\\4\end{pmatrix}(2x)^0(-3)^4$
$(2x-3)^4=16x^4-96x^3+216x^2-216x+81$

$(2)$ Find the co-efficient of the $x^3$ term in the expansion of $(2-5x)^5$ .

Remember $(x+y)^n=\Sigma_{i=0}^n \begin{pmatrix}n\\i\end{pmatrix}x^{n-i}y^i$ , the $x^3$ is when $i=3$
$\begin{pmatrix}5\\3\end{pmatrix}(2)^2(-5)^3=10\times 2\times -125=-5000$

$(3)$ Find the constant term in the expansion of $(x^2+\frac{3}{x^4})^6$

We need to find the term where the $x$ ‘s cancel out. Each term is $\begin{pmatrix}6\\i\end{pmatrix}(x^2)^{6-i}(\frac{3}{x^4})^i$ .
$\begin{pmatrix}6\\i\end{pmatrix}(x^{12-2i})(3^ix^{-4i})$ .
We need $12-2i-4i=0$ , hence $i=2$
Therefore, the co-efficient is $\begin{pmatrix}6\\2\end{pmatrix}\times3^2=135$

Leave a Comment

Filed under Algebra, Binomial Expansion Theorem, Counting Techniques, Year 11 Mathematical Methods

March 3, 2025 · 2:51 am

Derangements

From Wikipedia

In combinatorial mathematics, a derangement is a permutation of the elements of a set in which no element appears in its original position.

For example, if we have the set ${A, B, C}$ , there are $6$ permutations

$ABC, ACB, BAC, BCA, CAB, CBA$

But only $2$ of them are derangements – $BCA$ and $CAB$

I did a practical question on this here. In that question I used a tree diagram, but there must be a way to determine the number of derangements given $n$ elements.

We know $3$ elements has $2$ derangements, what about $4$ elements? We know there are $4!=24$ permutations

ABCD	BACD	CABD	DABC
ACBD	BADC	CADB	DACB
ACDB	BCAD	CBAD	DBAC
ABDC	BCDA	CBDA	DBCA
ADBC	BDAC	CDAB	DCAB
ADCB	BDCA	CDBA	DCBA

I have highlighted the derangements. when $n=4$ there are $9$ derangements.

Let’s try to generalise.

In how many ways can one element be in its original position?
$\begin{pmatrix}4\\1\end{pmatrix} \times 3!=24$
Choose one element from the 4 to be in its original position, and then arrange the remaining three elements.

So far, we have $D_3=4!-\begin{pmatrix}4\\1\end{pmatrix} \times 3!=0$

Clearly we are counting some arrangements multiple times, for example ABDC has A and B in the correct position, so we need to add all of the arrangements with $2$ elements in their original position.
In how many ways can two elements be in their original position?
$\begin{pmatrix}4\\2\end{pmatrix} \times 2!=12$

So now we have, $D_3=4!-\begin{pmatrix}4\\1\end{pmatrix} \times 3!+\begin{pmatrix}4\\2\end{pmatrix} \times 2!=12$

But once again we have counted some arrangements multiple times, so we need to subtract all of the arrangements with $3$ elements in their original position.
In how many ways can three elements be in their original position?
$\begin{pmatrix}4\\3\end{pmatrix} \times 1!=4$

So now we have, $D_3=4!-\begin{pmatrix}4\\1\end{pmatrix} \times 3!+\begin{pmatrix}4\\2\end{pmatrix} \times 2!-\begin{pmatrix}4\\3\end{pmatrix} \times 1!=8$

We now need to add all of the arrangements with $4$ elements in their original position.
In how many ways can four elements be in their original position?
$\begin{pmatrix}4\\4\end{pmatrix} \times 0!=1$

So now we have, $D_3=4!-\begin{pmatrix}4\\1\end{pmatrix} \times 3!+\begin{pmatrix}4\\2\end{pmatrix} \times 2!-\begin{pmatrix}4\\3\end{pmatrix} \times 1!+\begin{pmatrix}4\\4\end{pmatrix} \times 0!=9$

Hence,

$D_n=\begin{pmatrix}n\\0\end{pmatrix}\times n!-\begin{pmatrix}n\\1\end{pmatrix}\times (n-1)!+\begin{pmatrix}n\\2\end{pmatrix}\times (n-2)!-... \mp \begin{pmatrix}n\\n\end{pmatrix}\times 0!$

Which we can simplify to

$D_n=\Sigma_{i=0}^{n}(-1)^n\begin{pmatrix}n\\i\end{pmatrix}\times(n-i)!$

Leave a Comment

Filed under Counting Techniques, Interesting Mathematics, Year 11 Specialist Mathematics

February 25, 2025 · 9:28 am

Differentiating f(x)=e^x

We are going to differentiate $f(x)=e^x$ from first principals.

Remember the definition of a derivative is

(1) $\begin{equation*}f'(x)=\lim_{\limits{h\to 0}}\frac{f(x+h)-f(x)}{h}\end{equation*}$

If $f(x)=e^x$ , then

$\begin{equation*}f'(x)=\lim_{\limits{h \to 0}}\frac{e^{x+h}-e^x}{h}\end{equation}$

$\begin{equation*}f'(x)=\lim_{\limits{h \to 0}}\frac{e^x\times e^h-e^x}{h}\end{equation}$

$\begin{equation*}f'(x)=\lim_{\limits{h \to 0}}\frac{e^x(e^h-1)}{h}\end{equation}$

$\begin{equation*}f'(x)=e^x\lim_{\limits{h \to 0}}{\frac{e^h-1}{h}\end{equation}$

Let’s think about $\lim_{\limits{h \to 0}}\frac{e^h-1}{h}$

Remember $e$ is defined as $\lim_{\limits{n \to \infty}}(1+\frac{1}{n})^n$

(2) $\begin{equation*}\lim_{\limits{h \to 0}}{\frac{e^h-1}{h}\end{equation*}$

Let $y=e^h-1$ , as $h \to 0, e^h-1 \to 1-1=0$ hence $y \to 0$

If $y=e^h-1$ , then $h=ln(y+1)$

(3) $\begin{equation*}\lim_{\limits{y \to 0}}\frac{y}{ln(y+1)}\end{equation*}$

We are going to rewrite the equation $3$ as

(4) $\begin{equation*}\lim_{\limits{y \to 0}}\frac{\frac{1}{ln(y+1)}}{y}\end{equation*}$

And then we can write equation $4$ as

(5) $\begin{equation*}{\lim_{\limits{y \to 0}}{\frac{1}{\frac{1}{y}ln(y+1)}\end{equation*}$

Using log laws we can write equation $5$ as

(6) $\begin{equation*}\lim_{\limits{y \to 0}}\frac{1}{ln(y+1)^{\frac{1}{y}}}\end{equation*}$

Let $y=\frac{1}{n}$

As $y \to 0, n \to \infty$

equation $6$ becomes

(7) $\begin{equation*}\lim_{\limits{n \to \infty}}\frac{1}{ln(\frac{1}{n}+1)^n}\end{equation*}$

We can move the limit to inside the natural log

(8) $\begin{equation*}\frac{1}{ln(\lim_{\limits{n \to \infty}}(\frac{1}{n}+1)^n)}\end{equation*}$

And we know from the definition of $e$ that $e=\lim_{\limits{n \to \infty}}(1+\frac{1}{n})^n$

Hence, equation $8$ is

(9) $\begin{equation*}\frac{1}{ln(e)}=1\end{equation*}$

Back to our derivative

$\begin{equation*}f'(x)=e^x\lim_{\limits{h \to 0}}{\frac{e^h-1}{h}\end{equation}$

We know that $\lim_{\limits{h \to 0}}{\frac{e^h-1}{h}=1$ hence

$\begin{equation*}f'(x)=e^x\end{equation}$

Leave a Comment

Filed under Calculus, Differentiation, Year 12 Mathematical Methods

Tagged as differentiating e

February 19, 2025 · 6:15 am

Finding a Recursive Rule for a Sequence (Year 12 Mathematics Applications)

How do we go about finding the rule for a first order linear recurrence relation?

Something like

$\begin{equation*}5, 7, 11, 19, ...\end{equation}$

There isn’t a common difference (arithmetic sequence) or a common ratio (geometric sequence). Sometimes you can just see the rule, but an algorithm will be handy.

Let’s say our relationship is

(1) $\begin{equation*}T_{n+1}=bT_n+c, T_1=a\end{equation*}$

Referring back to our sequence $5, 7, 11, 19, ...$ , we know

(2) $\begin{equation*}7=b\times5+c\end{equation*}$

and

(3) $\begin{equation*}11=b\times7+c\end{equation*}$

We can solve equation $2$ and $3$ simultaneously

equation $2-$ equation $3$

$\begin{equation*}-4=-2b\end{equation}$

Hence $b=2$

Substitute $b=2$ into equation $2$

$\begin{equation*}7=2\times 5+c\end{equation}$

$\begin{equation*}7=10+c\end{equation}$

Hence $c=-3$

$\begin{equation*}T_{n+1}=2T_n-3, T_1=5\end{equation}$

Let’s try to generalise

If $T_{n+1}=bT_n+c, T_1=a$ , then

(4) $\begin{equation*}T_2=bT_1+c\end{equation*}$

and

(5) $\begin{equation*}T_3=bT_2+c\end{equation*}$

Equation $5 -$ equation $4$

$\begin{equation*}T_3-T_2=(T_2-T_1)b\end{equation}$

$\begin{equation*} b=\frac{T_3-T_2}{T_2-T_1}\end{equation}$

Hence, $b=\frac{T_{n+2}-T_{n+1}}{T_{n+1}-T_n}$

Once you know $b$ , substitute into either equation to find $C$ .

Example

Find the recursive rule for the following

$-8, -12, -20, -36, ...$

$\begin{equation*}b=\frac{T_{n+2}-T_{n+1}}{T_{n+1}-T_n}=\frac{-20-(-12)}{-12-(-8)}=\frac{-8}{-4}=2\end{equation}$

$\begin{equation*}-12=2\times-8+c\end{equation}$

$\begin{equation*}-12=-16+c\end{equation}$

Hence $c=4$ and $T_{n+1}=2T_n+4, T_1=-8$

It is also possible to find the rule using a Classpad (if it’s in the calculator section} by using an e-activity.

Leave a Comment

Filed under Sequences, Sequences, Year 12 Mathematics Applications

Tagged as first order linear sequences, recursive rules

Expression	Expansion	Co-efficients
$(x+y)^2$	$x^2+2xy+y^2$	$1, 2, 1$
$(x+y)^3$	$x^3+3x^2y+3xy^2+y^3$	$1, 3, 3, 1$
$(x+y)^4$	$x^4+4x^3y+6x^2y^2+4xy^3+y^4$	$1, 4, 6, 4, 1$

Deriving the Quadratic Equation formula

Logistic Growth Worked Example

Deriving the Logistic Growth Equation

Proving the Maximum Rate of Increase Happens When $P=\frac{k}{2}$

Intersecting Secant Theorem

Circle Geometry Question 2

Circle Geometry Question

Binomial Expansion Theorem

Worked Examples

Derangements

Differentiating f(x)=e^x

Finding a Recursive Rule for a Sequence (Year 12 Mathematics Applications)

Example

Recent Posts

Categories

Archives