A recursion rule

Sequences, grade 12 (4 units) Lesson 2 of 20Learning

A recursion rule

A recursion rule

A recursion rule says how each term comes from the term before it. The first term is given with it – that is where we start:

\begin{cases}a_1=3\\a_{n+1}=a_n+4\end{cases}

$\displaystyle a_{n+1}$ is the term after $\displaystyle a_n$ : the rule says that each term is 4 more than the term before it.
$\displaystyle {a_2}={a_1+4}={7}$ , $\displaystyle {a_3}={a_2+4}={11}$ , $\displaystyle {a_4}={a_3+4}={15}$ .
The rule holds for every natural $\displaystyle n$ , so the sequence goes on without end: $\displaystyle 3, 7, 11, 15, \ldots$

Example: a rule with a product

\begin{cases}a_1=2\\a_{n+1}=3a_n-1\end{cases}

We substitute in the rule, term after term:

$\displaystyle {n}={1}$ : $\displaystyle {a_2}={3a_1-1}={3\cdot2-1}={5}$ .
$\displaystyle {n}={2}$ : $\displaystyle {a_3}={3a_2-1}={3\cdot5-1}={14}$ .
$\displaystyle {n}={3}$ : $\displaystyle {a_4}={3a_3-1}={3\cdot14-1}={41}$ .

To reach $\displaystyle a_{10}$ we must pass through all the terms before it.

A formula by place and a recursion rule

The same sequence can have two definitions:

a formula by place gives each term straight from its place – $\displaystyle a_{100}$ too, at once;
a recursion rule gives each term from the term before it.

The sequence $\displaystyle 3, 7, 11, 15, \ldots$ is defined by the rule $\displaystyle {a_1}={3}$ , $\displaystyle {a_{n+1}}={a_n+4}$ , and also by the formula $\displaystyle {a_n}={4n-1}$ : $\displaystyle {a_1}={4\cdot1-1}={3}$ , $\displaystyle {a_2}={7}$ , $\displaystyle {a_3}={11}$ .

From a formula to a recursion rule

The example below finds a recursion rule for the sequence $\displaystyle {a_n}={4n-1}$ : we work out $\displaystyle a_{n+1}$ , and see how it differs from $\displaystyle a_n$ .

In place of $\displaystyle n$ we substitute $\displaystyle n+1$ : $\displaystyle {a_{n+1}}={4(n+1)-1}={4n+3}$ .
We subtract: $\displaystyle {a_{n+1}-a_n}={(4n+3)-(4n-1)}={4}$ .
Each term is 4 more than the one before it: $\displaystyle {a_{n+1}}={a_n+4}$ , and $\displaystyle {a_1}={4\cdot1-1}={3}$ .

Exercise 1

The sequence is defined by the recursion rule:

\begin{cases}a_1=5\\a_{n+1}=2a_n-3\end{cases}

Calculate:

a

\displaystyle a_2

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

b

\displaystyle a_3

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

c

\displaystyle a_4

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 2

The sequence is defined by the recursion rule:

\begin{cases}a_1=1\\a_{n+1}=a_n+2n\end{cases}

Calculate:

a

\displaystyle a_2

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

b

\displaystyle a_3

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

c

\displaystyle a_4

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 3

The sequence: $\displaystyle {a_n}={3n+2}$ .

a

Work out $\displaystyle a_{n+1}-a_n$ .

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

b

Write a recursion rule for the sequence, with its first term.

Exercise 4

A sequence is defined for every natural $\displaystyle n$ by the recursion rule:

\begin{cases}a_5=8\\n\cdot a_{n+1}=4a_n-3\end{cases}

a

Work out $\displaystyle a_6$ : substitute $\displaystyle {n}={5}$ .

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

b

Work out $\displaystyle a_4$ : substitute $\displaystyle {n}={4}$ , and solve the equation.

Formulas: sequences, grade 12 (4 units)

A sequence and the sum of its first terms

The term from the sum – $\displaystyle S_n$ the sum of the first $\displaystyle n$ terms
$\displaystyle S_n-S_{n-1}$ $\displaystyle =$ $\displaystyle a_n$ $\displaystyle n\geq2$
The first term is the sum of one term
$\displaystyle a_1$ $\displaystyle =$ $\displaystyle S_1$

The arithmetic sequence

The recursion rule – $\displaystyle d$ the common difference
$\displaystyle a_{n+1}=a_n+d$
The general term
$\displaystyle a_n$ $\displaystyle =$ $\displaystyle a_1+(n-1)d$
The arithmetic mean of the neighbours
$\displaystyle a_n$ $\displaystyle =$ $\displaystyle \frac{a_{n-1}+a_{n+1}}{2}$ $\displaystyle n\geq2$
The sum of the first $\displaystyle n$ terms
$\displaystyle S_n$ $\displaystyle =$ $\displaystyle \frac{n(a_1+a_n)}{2}$
The sum without the last term
$\displaystyle S_n$ $\displaystyle =$ $\displaystyle \frac{n(2a_1+(n-1)d)}{2}$

The geometric sequence

The recursion rule – $\displaystyle q$ the common ratio
$\displaystyle a_{n+1}=a_n\cdot q$
The general term
$\displaystyle a_n$ $\displaystyle =$ $\displaystyle a_1\cdot q^{n-1}$
The geometric mean: a term’s square is the product of its neighbours
$\displaystyle a_n^2$ $\displaystyle =$ $\displaystyle a_{n-1}\cdot a_{n+1}$ $\displaystyle n\geq2$
The sum of the first $\displaystyle n$ terms
$\displaystyle S_n$ $\displaystyle =$ $\displaystyle \frac{a_1(q^n-1)}{q-1}$ $\displaystyle q\neq1$

The infinite geometric sequence

The sum of the infinite sequence
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{a_1}{1-q}$ $\displaystyle -1<q<1,\ q\neq0$