Full Guide on Number Sequences And Series

Ramta solutions article on number patterns.

Introduction
Types of sequences
Sum of terms and ssigma notation
Series

Introduction

In mathematics, we define number patterns as patterns which contain a list of numbers that follow each other in a certain order. The order can vary from linear, quadratic, cubic, geometric, etc.

The Oxford English Dictinary definition of patterns is that a pattern is the regular arrrangement of lines, shapes, colours, etc. This is no different to the mathematical definition given in line 1.

You have seen what are patterns. Now let us introduce number sequences. A sequence is an ordered list of numbers. Sequences have patterns. Let us look at them:

\begin{align*} 1;2;3;4;5;...\\ 7;11;15;19;23;...\\ 1;4;9;15;24;...\\ 5;10;20;40;80;... \end{align*}

The above numbers are an example of sequences. The items on sequences are reffered to as terms. For example, for the first sequence, the 1 is the 1^st term, 2 is the 2^nd term, 3 is the 3^rd term, and so on. The ellipsis (…) means that the sequence continues forever.

Generally, sequences are in the following form:

$${\small T_1;T_2;T_3;T_3;T_4;T_5;\ldots}$$

Types of sequences

We will look at three types of sequences: Linear/arithmetic sequences, quadratic sequences and geometric sequences.

Arithmetic sequences

Arithmetic sequences have a common difference. Each consecutive term is an addition of this common difference. They resemble a straight line graph, which has the form, $y=mx+c$.

Given the sequence:

$${\small T_1;T_2;T_3;T_3;T_4;T_5;\ldots}$$

Assuming the sequence is linear, the common difference, $d$, is:

$${d=T_2-T_1=T_3-T_2=T_4-T_3=\ldots}$$

Because we have observed that the sequence is arithmetic, we can now determine its equation, which we call the general term, $T_n$.

$${T_n=a+(n-1)d}$$

Where $T_n$ is the term value (alternatively called the general term or the $n^{th}$ term), $a$ is the first term of the sequence, $n$ is the term number and $d$ is the common difference.

NB: $T_n$ will sometimes be referred to as the $n^{th}$ term.

Let us try a few examples

Find the general term:

$1;2;3;4;5;\ldots$
$5;10;15;20;25;\ldots$
$7;11;15;19;23;\dots$

Solutions

$\text{ }1;2;3;4;5;\ldots$
\begin{align*} \small a&=1 \text{ This is the first term}\\ \small d&=T_2-T_1\\ \small &=2-1\\ \small &=1 \end{align*}
Therefore the general term is:
\begin{align*} T_n&=a+(n-1)d\\ \small &=1+(n-1)(1)\\ \small &=1+n-1\\ \small &=n \end{align*}
$\text{ }5;10;15;20;25;\ldots$
\begin{align*} \small a&=5 \\ \small d&=T_2-T_1\\ \small &=10-5\\ \small &=5 \end{align*}
Therefore the general term is:
\begin{align*} \small T_n&=a+(n-1)d\\ \small &=5+(n-1)(5)\\ \small &=5+5n-5\\ \small &=5n \end{align*}
$\text{ }7;11;15;19;23;\dots$
\begin{align*} \small a&=7\\ \small d&=T_2-T_1\\ \small &=11-7\\ \small &=4 \end{align*}
Therefore the general term is:
\begin{align*} \small T_n&=a+(n-1)d\\ \small &=7+(n-1)(4)\\ \small &=7+4n-4\\ \small &=4n+3 \end{align*}

The examples above have shown us how to determine the general term. Now let us try a different example.

The third term of an arithmetic sequence is 17 and the ninth term is 41. Determine the values of $a$, $d$ and the $n^{th}$ term of this sequence.

Solutions

$${\small T_n=a+(n-1)d}$$

We are give the third term, $T_3$, and the ninth term, $T_9$ and their values. Now we must find the unknowns.

\begin{align} \small T_n&=a+(n-1)d\nonumber\\ \small T_3&=a+(3-1)d\nonumber\\ \small 17&=a+2d \tag{1} \end{align}

Doing the same for the ninth term.

\begin{align} \small T_n&=a+(n-1)d\\ \small T_9&=a+(9-1)d\\ \small 41&=a+8d \tag{2} \end{align}

Solve for $a$ and $d$ simultaneuously.

\begin{align} \small 17&=a+2d\tag{1}\\ \small 41&=a+8d \tag{2}\\ \small \text{(2)-(1): } 24&=6d \\ \small d&=4 \end{align}

Substituting $d$ onto $(1)$ or $(2)$, we get:

$${\small a=9}$$

The General term, $T_n$ can now be determined

\begin{align} \small T_n&=a+(n-1)d\\ \small T_n&=9+(n-1)(4)\\ \small T_n&=9+4n-4\\ \small T_n&=4n+5 \end{align}

Quadratic sequences

Quadratic sequences have the form of a quadratic graph, $y=ax^2+bx+c$. However, they only take positive intergers. Thus, they have the general term:

$${T_n=an^2+bn+c}$$

The quadratic sequence, has a common second difference. Consider the Fig. 1:

Procedure for deriving qudratic patterns' general term.

The first difference of a quadratic sequence yields a linear sequence. This prompts us to find the second difference which we can see from Fig. 1 that it is constant. When the second difference is constant, we can then conclude the sequence is quadratic.

We then need to solve the parameters, $a$, $b$ and $c$ by using the following equations:

\begin{align} \small T_1&=a+b+c\\ \small T_2-T_1&=3a+b\\ \small \text{Common }2^{nd}\text{ difference}&=2a \end{align}

Now let us take a few examples to make sense of this.

Find the general term of the following:

$\small 2;6;14;26;\dots$
$-1;0;3;8;\ldots$

Solutions

$\small \text{i. } 2;6;14;25;\dots$

\begin{gather*} \small \text{ }2;6;14;26;\dots\\ \small 4;8;12;\dots\\ \small 2;2;\ldots \end{gather*}

We can see that $T_1=a+b+c$ corresponds to 2 $T_2-T_1=3a+b$ corresponds to 4 and that common second diference is 4.

\begin{align} \small 2a&=4\\ \small a&=2\\ \\ \small 3a+b&=4\\ \small 3(2)+b&=4\\ \small 6+b&=4\\ \small b&=-2\\ \\ \small a+b+c=2\\ \small 2-2+c&=2\\ \small c&=2 \end{align}

The general term:

$${\small T_n=2n^2-2n+2}$$

$\small \text{ii. }-1;0;3;8;\ldots$

\begin{gather*} \small -1;0;3;8;\ldots\\ \small \text{ }1;3;5;\dots\\ \small 2;2;\ldots \end{gather*}

Here we can see that $T_1=a+b+c=$ corresponds to $-1$ $T_2-T_1=3a+b$ corresponds to 1 and that common second difference is is 2.

\begin{align} \small 2a&=2\\ \small a&=1\\ \\ \small 3a+b&=1\\ \small 3(1)+b&=1\\ \small 3+b&=1\\ \small b&=-2\\ \\ \small a+b+c=1\\ \small 1-2+c&=1\\ \small c&=0 \end{align}

The general term:

$${\small T_n=n^2-2n}$$

NB:You may also apply your knowledge of quadratic graphs to solve problems of quadratic patterns. This is more prevalent in the exam.

Geometric sequences

$${\small 2;6;18;54;\ldots}$$

Look at the sequence above. Every term, except the first, is the previous term multiplied by 3. Sequences like these are called geometric sequences.

This 3 is referred to as the common ratio. which we find by:

$\frac{6}{2}=\frac{18}{6}=\frac{54}{18}=3$

Let us now try to understand how we made this sequence.

$2$
$6=2\times3^1$
$18=6\times3=2\times3\times3=2\times3^2$
$54=18\times3=2\times3\times3\times3=2\times3^3$

We can see that for the second term and beyond, the sequence is made by multiplying the previous term with 3. This creates a pattern of base 3 powers. Therefore the general term is:

$${T_n=2\times3^{n-1}}$$

This 3 is the common ratio
The $n-1$ accounts for the fact that the 3 does not appear on $T_1$ (or rather say, it is $3^0$).

Generally, the general term of a geometric sequences is:

$${T_n=a{\cdot}r^{n-1}}$$

Where:

$T_n$ is the general term
$a$ is the first term.
$r$ is the common ratio
$n$ is the term number

To find $r$:

$${\small r=\frac{T_2}{T_1}=\frac{T_3}{T_2}=\frac{T_4}{T_3}=\ldots}$$

Now let us try some few examples.

Find the general term of the following:

$128;64;32;\ldots$
$0,25;0,5;1;\ldots$

Solutions

$\text{i. }128;64;32;\ldots$

$a=128$
$r=\frac{T_2}{T_1}=\frac{64}{128}=\frac{1}{2}=2^{-1}$

The general term is:

\begin{align} \small T_n&=a{\cdot}r^{n-1}\\ \small &=128{\cdot}(2^{-1})^{n-1}\\ \small &=128{\cdot}2^{-n+1} \end{align}

$\text{ii. }0,25;0,5;1;\ldots$

$\small a=0,25$

$\small r=\frac{T_2} {T_1}=\frac{0,5}{0,25}=2$

The general term is:

\begin{align} \small T_n&=a{\cdot}r^{n-1}\\ \small &=0,25{\cdot}2^{n-1}\\ \small &=\frac{1}{4}{\cdot}2^{n-1} \end{align}

Now let us try one last example

The second term of geometric term is 3, while the seventh is $22\frac{25}{32}$. Calculate the $a$, $r$, and the general term of this sequence.

Solutions

$${T_n=a{\cdot}r^{n-1}}$$

$T_2=3$

$T_7=22\frac{25}{32}=\frac{729}{32}$

\begin{align} \small T_2&=a{\cdot}r^{2-1}\\ \small 3&=ar^1 \tag{1}\\ \\ \small T_7&=a{\cdot}r^{7-1}\\ \small \frac{729}{32}&=ar^6 \tag{2} \end{align}

Dividing (2) with (1), we get:

\begin{align} \small \frac{729}{32\times3}&=\frac{ar^6}{ar^1}\\ \small \frac{243}{32}&=r^5\\ \small r&=\sqrt[5]{\frac{243}{32}}\\ \small r&=\frac{3}{2} \end{align}

Sum of terms and sigma notation

We have looked at three types of sequences. Now let us take it a notch up and look at their sum. But first, let us begin with an introduction.

You are building a pattern of shapes using sticks and realised the shapes have the following sequence:

$${\small 1;4;7;10;13;17\ldots}$$

How many sticks would yoou need to build these your patterns? Say a pattern of six shapes?

To do this, you will have to add the sticks for all six shapes. That is:

$${\small 1+4+7+10+13+17}$$

The sum for all six is 53.

Calculate the sum of the following:

$\small 3+4+5+6$

The answer is $18$

$\small 1+4+9+16+25$

The answer is $55$

$\small 1+2+4+8+16$

The answer is $31$

Imagine if we were to add 50 numbers. They would fill this page or you would stop the task. Now let us introduct a compact way of writing sums.

Sigma notation

We use the sigma notation to write a sum of numbers in a short form.

Remember that to write the sum of all 50 or 100 terms would be tedious. Now let us explore this sigma notation.

Say we are given this sum: $\small 15+30+35+40 \ldots$ up to 50 terms. We would stop the calculation or run to EXCEL, right? Yes.

We will write the sum of terms as $S_n$, where $n$ is the term number. The sigma notation is written as:

$${S_n=\sum_{n=1}^{m}T_n}$$

where:
$n=1$ is the start index
$m$ is the end index and
$T_n$ is your general term

The start index tells you where you should start calculating, and the end index tells where you should end your calculation.

Let's try some examples

Write the following in sigma notation:

$\small 1+2+3+4+\ldots$ up to 20 terms
$\small 25+30+35+40+45+50+55$
$\small 40+80+160+320$
$\small 3+3+3+3+3+3$

Solutions

$\small \text{i. }1+2+3+4+\ldots$

This looks like an arithmetic sequence.

Its general term is $T_n=n$
$T_1=1$ so start index is $n=1$
We are told "up to 20 terms". So the end index is 20.
Counting from 1 to 20, we have 20 terms, therefore $n=20$.

The sigma notation is:

$${\small S_{20}=\sum_{n=1}^{20}n}$$

$\small \text{ii. }25+30+35+40+45+50+55$

General term: $T_n=5n+20$
Start index is 1
There are six terms, so we will say end index is 7
$n=7$. There are 7 terms

It can also be written as:

This considers that the pattern is a multiple of 5 and start counting at $n=5$ and ends at 11.

$\small \text{iii. }40+80+160+320$

This is a geometric sequence. $a=40$
$\small r=\frac{80}{40}=2$
There are 4 terms.

Alternatively,

$\small \text{iv. } 3+3+3+3+3+3$

This is a sum of constant terms. There is no general solution. Therefore the sigma notion is:

$${\small S_6=\sum_{n=1}^{6}3}$$

To know number of terms when given a sigma notation, use the following:

$${\small \text{Number of terms}=End -\text{ Start}+1}$$

Write the first four terms of the sum, and the the number of terms of the following:

$\small \sum_{n=1}^{5}2n$
$\small \sum_{n=3}^{7}2an$
$\small \sum_{n=3}^{12}6$
$\small \sum_{n=1}^{b}n^2$

Solutions

$\small \text{i. } \sum_{n=1}^{5}2n$

The first four terms:

\begin{align} \small &=2(1)+2(2)+2(3)+2(4)\\ \small &=2+4+6+8 \end{align}

The number of terms:

\begin{align} \small \text{Number of terms}&=End -\text{ Start}+1\\ \small &=5-1+1\\ \small &=5 \end{align}

$\small \text{ii. } \sum_{n=3}^{7}2an$

The first four terms:

\begin{align} \small &=2a(3)+2a(4)+2a(5)+2a(6)\\ \small &=6a+8a+10a+12a \end{align}

The number of terms:

\begin{align} \small \text{Number of terms}&=End -\text{ Start}+1\\ \small &=7-3+1\\ \small &=5 \end{align}

$\small \text{iii. }\sum_{n=3}^{12}6$

The first four terms:

\begin{align} \small &=6+6+6+6 \end{align}

The number of terms:

\begin{align} \small \text{Number of terms}&=End -\text{ Start}+1\\ \small &=12-3+1\\ \small &=10 \end{align}

$\small \text{iv. }\sum_{n=1}^{b}n^2$

The first four terms:

\begin{align} &=1^2+2^2+3^2+4^2\\ &=1+4+9+16 \end{align}

The numberof terms:

\begin{align} \small \text{Number of terms}&=End -\text{ Start}+1\\ \small &=b-1+1\\ \small &=b \end{align}

Series

We have seen how to write sum of numbers. Now we will introduce a way of calculating the sum of numbers.

A series is a sum of terms of a sequence. Mathematically, we represent a series as:

$${\small S_n=T_1+T_2+T_3+T_4+\ldots}$$

In grade 12, we look at finite arithmetic series, finite geometric series and the infinite series.

Finite arithmetic series

To calculate the sum of terms of an arithmetic sequence, you use the following equations:

\begin{align} \small S_n&=\frac{n}{2}(a+l)\tag{1}\\ \small S_n&=\frac{n}{2}[2a+(n-1)d]\tag{2} \end{align}

where:

$\small S_n$ is the sum of terms
$\small n$ is the term number
$\small a$ is the first term
$\small l$ is the last term of the sequence/series
$\small d$ is the common first difference.

If you are given the first terms and other successive terms, use (1). If given the first and the last tersm, use (2).

Now let us try a few examples to understand these formulas.

Find the sum of the first 25 terms:

$\small 1+2+3+4\ldots$
$\small -8-5-2+1+\ldots$
$\small 22+17+12+7+\ldots$

Solutions

$\small \text{i. }1+2+3+4\ldots$

$\small a=1$
$\small d=T_2-T_1=2-1=1$
$n=25$
$\small S_{25}=?$

\begin{align} \small S_n&=\frac{n}{2}[2a+(n-1)d]\\ \small S_{25}&=\frac{25}{2}[2(1)+(25-1)(1)]\\ \small &=325 \end{align}

$\small \text{ii. } -8-5-2+1+\ldots$

$\small a=-8$
$\small d=T_2-T_1=-5-(-8)=3$
$n=25$
$\small S_{25}=?$

\begin{align} \small S_n&=\frac{n}{2}[2a+(n-1)d]\\ \small S_{25}&=\frac{25}{2}[2(-8)+(25-1)(3)]\\ \small &=700 \end{align}

$\small \text{iii. }22+17+12+7+\ldots$

$\small a=-8$
$\small d=T_2-T_1=17-22=-5$
$n=25$
$\small S_{25}=?$

\begin{align} \small S_n&=\frac{n}{2}[2a+(n-1)d]\\ \small S_{25}&=\frac{25}{2}[2(22)+(25-1)(-5)]\\ \small &=-950 \end{align}

Now we will try a different example that requires some algebra.

The sum of the first 16 terms of an arithmetic sequence is 528. Determine the first term and the common first difference if the seventh term is 21.

Solutions

$\small S_{16}=528$
$\small T_7=27$
$\small a=?$
$\small d=?$

\begin{align} \small S_n&=\frac{n}{2}[2a+(n-1)d]\\ \small S_{16}&=\frac{16}{2}[2a+(16-1)]\\ \small 528&=8(2a+15d)\\ \small 66&=2a+15d\tag{1} \end{align} \begin{align} \small T_n&=a+(n-1)d\\ \small T_7&=a+(7-1)d\\ \small 27&=a+6d\tag{2} \end{align}

Solving for $a$ and $d$ simultaneously, we get:

$${a=3 \text{ and }d=4}$$

Finite geometric series

Turning our attention to the sum of terms of the geometric sequence, the sum of terms can be calculated as:

\begin{align} \small S_n=\frac{a(r^n-1)}{r-1}\text{, }r\gt1\tag{3}\\ \small S_n=\frac{a(1-r^n)}{1-r}\text{, }r\lt1\tag{4} \end{align}

The use of equations (3) and (4) depend on the value of r, If $\small r\gt1$, we use (3) and if $\small r\lt1$, we use (4).

Now let us try a few examples

Find the sum of the first 12 terms:

$\small 1+3+9+27\ldots$
$\small 9+3+1+\frac{1}{3}+\ldots$
$\small 120+60+30+15+\ldots$

Solutions

$\small \text{i. }1+3+9+27\ldots$

$\small a=1$
$\small d=\frac{T_2}{T_1}=\frac{3}{1}=3$
$n=12$
$\small S_{12}=?$

\begin{align} \small S_n&=\frac{a(r^n-1)}{r-1}\\ \small S_{12}&=\frac{1(3^{12}-1)}{3-1}\\ \small &=\text{265 720} \end{align}

$\small \text{ii. }9+3+1+\frac{1}{3}+\ldots$

$\small a=9$
$\small d=\frac{T_2}{T_1}=\frac{3}{9}=\frac{1}{3}$
$n=12$
$\small S_{12}=?$

\begin{align} \small S_n&=\frac{a(r^n-1)}{r-1}\\ \small S_{12}&=\frac{9[(\frac{1}{3})^{12}-1]}{\frac{1}{3}-1}\\ \small &=\text{13,50} \end{align}

$\small \text{iii. }120+60+30+15+\ldots$

$\small a=120$
$\small d=\frac{T_2}{T_1}=\frac{60}{120}=\frac{1}{2}$
$n=12$
$\small S_{12}=?$

\begin{align} \small S_n&=\frac{a(r^n-1)}{r-1}\\ \small S_{12}&=\frac{120[(\frac{1}{2})^{12}-1]}{\frac{1}{2}-1}\\ \small &=\text{239,94} \end{align}

Infinite series

Consider what would happen if we sum the following:

$\small 1+2+3+4+\ldots$
$\small 5+10+20+40+\ldots$
$9+3+1+\frac{1}{3}+\ldots$

The first two series will always return a value which is different per any value of $\small n$. the same cannoy be saodd about the last series. Series like the first two are termed, the finite series, while the third is termed the infinite series.

Series are classified as convergent or divergent.

Divergent series are those that do not get closer a certain number as the number of terms increase.

Convergent series are those that do not get closer to a certain number as the number of terms increase.

NB: All arithmetic series are divergent, and not all geometric series are divergent.

For a geonetric series to converge, $\small -1{\lt}r{\lt}1$. This means $\small r$ must be greater than $\small -1$ or less than 1.

Series that converge, sum up to infinity. The sum to infinity is:

$${\small S_{\infty}=\frac{a}{1-r}}$$

Find the sum to infinity:

$\small \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+$
$\small 9+3+1+\frac{1}{3}+\ldots$

Solutions

$\small \text{i. }\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots$

First, let us check if this series converges.

$\small r=\frac{T_2}{T_1}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}$
The series converges. Therefore, we can use sum to infinity.

$\small \text{i. }\small 9+3+1+\frac{1}{3}+\ldots$

First, let us check if this series converges.

$\small r=\frac{T_2}{T_1}=\frac{3}{9}=\frac{1}{3}$
The series converges. Therefore, we can use sum to infinity.

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Full Guide on Number Sequences And Series

Contents

Introduction

Types of sequences

Arithmetic sequences

Quadratic sequences

Geometric sequences

Sum of terms and sigma notation

Sigma notation

Series

Finite arithmetic series

Finite geometric series

Infinite series