3.3: Recognising Sequences

For some of the sequences, we can predict the $n^{th}$ term easily, and the explicit (general) formula can be checked using induction. In this section, we will explore this kind of sequence.

Example $\PageIndex{1}$: Harmonic sequence

Find a formula for the $n^{th}$ term of the a sequence that has the following initial terms: $\displaystyle \frac{1}{2}, \displaystyle\frac{1}{3}, \displaystyle\frac{1}{4}, \displaystyle\frac{1}{5}, \displaystyle\frac{1}{6}, \cdots$.

Answer

$\displaystyle\frac{1}{n+1}$

Caution

Two sequences may start with the same initial terms but diverge later on.

Example $\PageIndex{2}$: One's

$1 \times 11=11$

$11 \times 11=121$

$111 \times 111=12321$

$1111 \times 1111=1234321$

Can you predict the pattern?

Example $\PageIndex{3}$: Perfect Squares

Find a formula for the $n^{th}$ term of the following sequence $1, 4, 9, 16, 25,\cdots$.

Answer

$n^2$.

Here is another way to find the $n^{th}$ term:

Note that the difference in the sequence $1, 4, 9, 16, 25,\cdots$ is $3, 5, 7, 9,\cdots$ and the difference in this sequence is $2$. Therefore the $n^{th}$ can be written as a quadratic formulae $pn^2+qn+c$. $p, q, c$ can be found using the following information:

Table $\PageIndex{3}$: Perfect Squares

	$t_n$	$d_n=t_n-t_{(n-1)}$	$d=d_n-d_{(n-1)}$
$n=1$	$p+q+c$
$n=2$	$4p+2q+c$	$3p+q$
$n=3$	$9p+3q+c$	$5p+q$	$2p$
$n=4$	$16p+4q+c$	$7p+q$	$2p$

Hence, in our case, $2p=2$. Therefore, $p=1$. Now $3p+q=3$ and $p=2$. Thus $q=0$. Now $p+q+c=1$, which implies $c=0$. Hence, $t_n=n^2$.

Quadratic Sequences:

A sequence is called a quadratic sequence if the differences of consecutive terms, $d_n=t_n−t_{(n−1)}$ differ by the same amount $d=d_n−d_{(n−1)}, \forall n \in \bf{N}$. In this case, the $n^{th}$ term of the sequence is given by

$t_n= a+(n-1) d_1+(n-1)(n-2)\frac{d}{2}$,

where $d_1$ is the first difference between the first and second terms of the sequence, and $d$ is the common second difference. This result can be shown by using induction. We will explore this in the next section.

Below is another way to solve this:

Example $\PageIndex{4}$: Quadratic Sequences

Find a formula for the $n^{th}$ term of the following sequence: $2, 6, 12, 20, \cdots$

$2 \, 6\, 12\, 20$

$4 \,6 \,8$

$2\, 2$

Hence the $n^{th}$ term has a term $(2/2) n^2.$

Table $\PageIndex{4}$: Quadratic Sequences

$t_n$	$2$	$6$	$12$	$20$
$n^2$	$1$	$4$	$9$	$16$
$n^2-t_n$	$1$	$2$	$3$	$4$

Notice that $n^2-t_n$ is an arithmetic sequence with the first term $1$ and the common difference is $1$. Thus $n^2-t_n =1+1(n-1)$. Hence $t_n=n^2-n.$

Triangular numbers

Thinking Out Loud:

$1, 3, 6, 10, 15, 21, 28, ...$ are triangular numbers: you can arrange them in a triangular array:

$T_n = n$^th triangular number. What is the hundredth triangular number?

Example $\PageIndex{5}$: Hexagonal Tilling (Centered hexagonal numbers)

Find the $n^{th}$ term of the sequence $1, 7, 19, 37, \cdots$. There are 6 triangular numbers and 1 center; therefore, $6 \dfrac{(n-1)n}{2}+1=3n(n-1)+1=3n^2-3n+1.$

$6\dfrac{(n-1)n}{2} +1=3(n(n-1))+1=3n^2-3n+1.$

 

Example $\PageIndex{6}$:

Consider the sequence of tilling using hexagons. The hexagon numbers are the sequence $1, 6, 15, · · ·$. Predict the nth term. Explain your prediction.

Table $\PageIndex{6}$: Hexagon numbers

$A$	Term # ($n$)	$1$	$2$	$3$	$4$
$B$	# of hexagons	$1$	$6$	$15$	$28$
$C$	$\dfrac{B}{A}$	$1$	$3$	$5$	$7$
$D$	$2(n)-1$	$1$	$3$	$5$	$7$

 

Since we know that $(A)(C) = B$, we can conclude that the term number $(A)=n,$ multiplied by $C (2n-1)$ will give you the corresponding number in the sequence $B$.
Therefore: $$t_n = n(2n-1)= 2n^2-n.$$

Tower of Hanoi

According to the legend of the Tower of Hanoi (formerly the "Tower of Brahma" in a temple in the Indian city of Benares), the temple priests are to transfer a tower consisting of 64 fragile disks of gold from one part of the temple to another, one disk at a time. The disks are arranged in order, no two of them the same size, with the largest on the bottom and the smallest on top. Because of their fragility, a larger disk may never be placed on a smaller one, and there is only one intermediate location where disks can be temporarily placed. It is said that before the priests complete their task the temple will crumble into dust and the world will vanish in a clap of thunder.

Let A, B and C be the posts. Then •1 disk: 1 move

Move 1: move disk 1 to post C

•2 disks: 3 moves

Move 1: move disk 2 to post B
Move 2: move disk 1 to post C
Move 3: move disk 2 to post C

•3 disks: 7 moves

Move 1: move disk 3 to post C
Move 2: move disk 2 to post B
Move 3: move disk 3 to post B
Move 4: move disk 1 to post C
Move 5: move disk 3 to post A
Move 6: move disk 2 to post C
Move 7: move disk 3 to post C

The number of moves needed to transfer n disks from post-A to post C is $2M+1$, where $M$ is the number of moves needed to transfer $n-1$ disks from post A to post C. This is called a recursive sequence. Can we able to guess a formula depending on n only?

Table $\PageIndex{7}$: Tower of Hanoi

Number of Disks	Min. number of Moves
$1$	$1$
$2$	$3$
$3$	$7$
$4$	$15$
$5$	$31$

 

From this pattern, we can guess the formula for finding the minimum number of moves it takes to transfer n disks from post-A to post C is: $2^n - 1.$ This guess can be proved by using induction.

Recursive Sequences

Definition

A recurrence relation for a sequence $\{a_n\}$ is formula that relates to each term $a_n$ to its predecessors $a_0,a_1, \cdots, a_{n-1}$.

Thinking Out Loud:

A tiling covers a region using tile pieces from some given set to cover the region without overlaps. How many ways can you arrange the $2 \times 1$ dominoes to cover the $2 \times n$ checkerboard?

Fibonacci Sequences

Fibonacci sequence: $1, 1, 2, 3, 5, 8, 13, 21, 34, 55, , ...$

Let $F_n = n$^$th$ Fibonacci Number.The $n$ term in the Fibonacci sequence is obtained by adding the previous two terms.

That is, $F_n = F$_{$(n - 1)$} $+ \, F$_{$(n - 2)$}, $n > 2, \, F_1=1,$ and $F_2 = 1$

Some facts about the Fibonacci sequence :

• The sum of the first n even-numbered Fibonacci numbers is one less than the next Fibonacci number.
• The only square Fibonacci numbers are $0, 1$ and $144.$
• The sum of the first n odd-numbered Fibonacci numbers is the next Fibonacci number.

Example $\PageIndex{9}$:

Consider the Fibonacci sequence: $1, 1, 2, 3, 5, 8, 13, 21, 34, 55, , ...$

The squares of the Fibonacci sequence: $1, 1 ,4,9, 25, 64, 169 ....$. Consider the sum of the squares of consecutive terms, F²_{$(n - 1)$} $+ \, F^2$_{$(n - 2)$}, for $n >2$. That is:

$1+1=2$

$1+4 =5$

$4+9=13$

What can you say about the resulting number?

Answer

It is Fibonacci. In fact, $F_{(2n-3)} = F^2$_{$(n - 1)$} $+ \, F^2$_{$(n - 2)$.}

Example $\PageIndex{11}$: More Fibonacci

Let $a_n$ be the Fibonacci sequence. That is $a_n=a_{n-1}+a_{n-2}, n \in \mathbb{N}, \mbox{ with } a_1=1,a_0=0$, where $\mathbb{N}$ be the set of all natural numbers.
Consider the Fibonacci squares illustrated in the figure:

Let $h_n$ be the shortest (perpendicular) distance between $n^{th}$ parallel diagonal vectors. Consider the figure:

Comparing the  area of trapezoids, we get

$h_n=\displaystyle\frac{a_n^2-a_{n-1}^2} {\sqrt{2}(a_n-a_{n-1})} =\displaystyle\frac{1}{\sqrt{2}}\left(a_n+a_{n-1}\right).$

Hence, $h_n$ is a Fibonacci-like sequence.

 

• Tower of Hanoi by https://texample.net/tikz/examples/towers-of-hanoi/