Tag Archives: recursive rules

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.

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