3.2: Arithmetic Sequences, Geometric Sequences : Visual Reasoning, and Proof by Induction

Arithmetic Sequences

Definition

Arithmetic sequences are patterns of numbers that increase (or decrease) by a set amount each time when you advance to a new term. You can determine the next term by adding the difference between any two terms to the final one to generate the next term. Let $a$ be the initial term and $d$ be the difference, then the $n^{th}$ term of the arithmetic sequence can be expressed as $t_n = a + (n - 1)d$.

Example $\PageIndex{1}$:

Table $\PageIndex{1}$: arithmetic sequence

$n$	$n^{th}$ Term $=t_n$	$t_n - t_{(n - 1)}$
5	11	2
6	13	2
7	15	2
8	17	2

As you can see, this sequence's terms increase by $2$ each time.

Example $\PageIndex{2}$:

Given the following sequence, can we determine $a$ and $d$ and give the sequence's general form?

$1, 4, 7, 10, 13, 16, 19...$

So:

• $a = 1$, because that is the first term in the sequence.
• $d = 3$, because the terms increase by 3 each time.

So the general form for the sequence is:

$t_n = 1 + (n - 1)3=3n+2$

Thinking Out Loud:

Is a sequence still an arithmetic sequence if the difference changes with each iteration, even if it is still added? Why or why not? What truly defines an arithmetic sequence?

Finite Sum of Arithmetic Sequences

There are two equivalent formulas for determining the finite sum of an arithmetic sequence. Here, we shall derive both the formulas and show how they are equal.

Example $\PageIndex{3}$:

Consider $S_n = a \,\,+ (a+d) \,\,+ \cdots +(a+(n-2)d)+(a+(n-1)d)$

Now, $S_n = (a+(n-1)d)+(a+(n-2)d+ \cdots+ (a+d)\,+a$.

Then, $2S_n= n(a+ a+(n-1)d) \,= n(2a+(n-1)d)$.

Hence, $S_n=\displaystyle \frac {n}{2}(t_1+ t_n) \,= \displaystyle \frac {n}{2}(2a+(n-1)d)$.

Let's illustrate these formulas by using the sequence $t_n = 5 + (n - 1)2= 2n+3$.

Example $\PageIndex{4}$:

The first formula we can use looks like this:

$S_n = n \left(\displaystyle \frac{a_1 + a_n}{2} \right)$

As we can see, this formula takes the average between the first and last terms, and multiplies by the number of terms in the series. So, if we use our series $t_n = 5 + (n - 1)2$ and we want the sum for the first 15 terms, our calculation will look like this:

$S-{15} = 15 \left( \frac{5 + 33}{2} \right)$

$= 15 \left( \frac{38}{2} \right)$

$= 15(19)$

$= 285$

Example $\PageIndex{5}$:

The second formula we can use looks like:

$S_n = \displaystyle \frac{n}{2} \left(2a + (n - 1)d \right)$

As we can see, this method doesn't need us to know the value of the $n$^$th$ term, just which term it is. Using our series $t_n = 5 + (n - 1)2$, our calculations look like this when we are looking for the sum of the first 15 terms:

$S = \displaystyle \frac{15}{2} \left(2(5) + ((15) - 1)(2) \right)$

$= \displaystyle \frac{15}{2} \left(10 + (14)(2) \right)$

$= \displaystyle \frac{15}{2} \left( 38 \right)$

$= 285$

So these two methods look to be equivalent so far. Let's show that this is true in the general case:

Example $\PageIndex{6}$:

Consider $S = \displaystyle \frac{n}{2} \left(2a + (n - 1)d \right)$, where $n$ is the number of the term that is the endpoint, $a$ is the series' starting value, and $d$ is the difference between any two consecutive terms.

Consider $S = n \left(\displaystyle \frac{a_1 + a_n}{2} \right)$

Since $a_n = t_n = a + (n - 1)d$,

And $a_1 = a$,

Then $S = n \left(\displaystyle \frac{a + [a + (n - 1)d]}{2} \right)$

$= n \left(\displaystyle \frac{2a + (n - 1)d}{2}\right)$

$= \displaystyle \frac{n}{2} \left(2a + (n - 1)d\right)$

Let's explore summation notation, which will be useful to represent finite sums:

Sigma (Summation) Notation

A finite sum requires adding up long strings of numbers. To make writing down these lengthy sums easier, we look at some new notation here, called sigma notation (also known as summation notation). The Greek capital letter $Σ$, sigma, is used to express long sums of values in a compact form. For example, if we want to add all the integers from 1 to 20 without sigma notation, we have to write

$$1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20.$$

We could probably skip writing a couple of terms and write

$$1+2+3+4+⋯+19+20,$$

which is better, but still cumbersome. With sigma notation, we write this sum as

$$\sum_{i=1}^{20}i$$

which is much more compact. Typically, sigma notation is presented in the form

$$\sum_{i=1}^{n}a_i$$

where $a_i$ describes the terms to be added, and i is called the $index$. Each term is evaluated, then we sum all the values, beginning with the value when $i=1$ and ending with the value when $i=n.$ For example, an expression like $\sum_{i=2}^{7}s_i$ is interpreted as $s_2+s_3+s_4+s_5+s_6+s_7$. Note that the index is used only to keep track of the terms to be added; it does not factor into the calculation of the sum itself. The index is therefore called a dummy variable. We can use any letter we like for the index. Typically, mathematicians use i, j, k, m, and n for indices.

Let’s try a couple of examples of using sigma notation.

Example $\PageIndex{7}$: Using Sigma Notation

 

1. Write in sigma notation and evaluate the sum of terms $3^i$ for $i=1,2,3,4,5.$
2. Write the sum in sigma notation:

$$1+\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+\dfrac{1}{25}.$$

Solution

1. Write $$\sum_{i=1}^{5}3^i=3+3^2+3^3+3^4+3^5=363.$$
2. The denominator of each term is a perfect square. Using sigma notation, this sum can be written as $\sum_{i=1}^5\dfrac{1}{i^2}$.

Exercise $\PageIndex{1}$

 

Write in sigma notation and evaluate the sum of terms $2^i$ for $i=3,4,5,6.$

Hint

Use the solving steps in Example as a guide.

Answer

$\sum_{i=3}^{6}2^i=2^3+2^4+2^5+2^6=120$

The properties associated with the summation process are given in the following rule.

Rule: Properties of Sigma Notation

Let $a_1,a_2,…,a_n$ and $b_1,b_2,…,b_n$ represent two sequences of terms and let c be a constant. The following properties hold for all positive integers n and for integers m, with $1≤m≤n.$

1. $\sum_{i=1}^nc=nc$
2. $\sum_{i=1}^n ca_i=c\sum_{i=1}^na_i$
3. $\sum_{i=1}^n(a_i+b_i)=\sum_{i=1}^na_i+\sum_{i=1}^nb_i$
4. $\sum_{i=1}^n(a_i−b_i)=\sum_{i=1}^na_i−\sum_{i=1}^nb_i$
5. $\sum_{i=1}^na_i=\sum_{i=1}^ma_i+\sum_{i=m+1}^na_i$

Proof:

We prove properties 2. and 3. here, and leave proof of the other properties to the Exercises.

2. We have

$\sum_{i=1}^nca_i=ca_1+ca_2+ca_3+⋯+ca_n=c(a_1+a_2+a_3+⋯+a_n)=c\sum_{i=1}^na_i$.

3. We have

$$\begin{align} \sum_{i=1}^{n}(a_i+b_i)&=(a_1+b_1)+(a_2+b_2)+(a_3+b_3)+⋯+(a_n+b_n) \\ &=(a_1+a_2+a_3+⋯+a_n)+(b_1+b_2+b_3+⋯+b_n) \\ & =\sum_{i=1}^na_i+\sum_{i=1}^nb_i. \end {align}$$□

A few more formulas for frequently found functions simplify the summation process further. These are shown in the next rule, for sums and powers of integers, and we will explore further in later examples.

Rule: Sums and Powers of Integers

1. The sum of n integers is given by

$$\sum_{i=1}^ni=1+2+⋯+n=\dfrac{n(n+1)}{2}.$$

2. The sum of consecutive integers squared is given by

$$\sum_{i=1}^ni^2=1^2+2^2+⋯+n^2=\dfrac{n(n+1)(2n+1)}{6}.$$

3. The sum of consecutive integers cubed is given by

$$\sum_{i=1}^ni^3=1^3+2^3+⋯+n^3=\dfrac{n^2(n+1)^2}{4}.$$

Proof

We leave proof (by induction) of the rules to the Exercises.

Geometric Sequences

Definition: 

Geometric sequences are patterns of numbers that increase (or decrease) by a set ratio with each iteration. You can determine the ratio by dividing a term by the preceding one. Let $a$ be the initial term and $r$ be the ratio, then the $n$^$th$ term of a geometric sequence can be expressed as $t_n = ar$ ^{$(n - 1)$}.

Example $\PageIndex{8}$

Table $\PageIndex{8}$: Geometric sequence

$n$	$t_n$	$t_n / t$ $n - 1$
1	3
2	6	2
3	12	2
4	24	2

So we can see that $r = 2$ since the ratio between any two consecutive terms is $2$.

Example $\PageIndex{9}$

Given the sequence $-3, 6, -12, 24, -48...$, can we:

1. Determine $a$ and $r$
2. Express the general form of the sequence

So:

• $a = -3$ because that is the sequence's initial term.
• $r = -2$, because if we divide any term by the preceding one, that is the result.

So the general form for the sequence is:

$t_n = -3(-2)$ ^{$(n - 1)$}.

Finite Sum of Geometric Sequences

Let's use the Gauss method to find a general case for the sum of a geometric sequence:

Example $\PageIndex{10}$

Let $r \ne 1.$

Consider $S_n = a + ar + ar^2 + ar^3 + ... + ar$ ^{$(n - 1)$}

Now, $(1)S_n = \, a \, + ar + ar^2 + ar^3 + ... + ar$ ^{$(n - 1)$}

$- (r)S_n = ar + ar^2 + ar^3 + ... + ar$ ^{$(n - 1)$} $+ ar^n$

$\left(1 - r\right)S_n = a - ar^n$

$S_n = \displaystyle \frac{a(1 - r^n)}{1 - r}$

That is, $$\sum_{k=0}^{(n-1)} a r^k = \displaystyle \frac{a(1 - r^n)}{1 - r}, r \ne 1.$$

Sum of Integers

Observe:

$1 = 1$

$1 + 2 = 3$

$1 + 2 + 3 = 6$

$1 + 2 + 3 + 4 = 10$

$1 + 2 + 3 + 4 + 5 = 15$

$1 + 2 + 3 + ... + n = ?$ This is the finite sum of first $n$ positive integers. Below, we have shown two ways of finding this sum:

Example $\PageIndex{11}$

Let's figure out a general case for a sum of integers beginning with $1$ and ending with $n$ using Gauss' method:

Let $S_n = 1 \, + \, \, \, \, \, \, \, 2 \, \, \, \, \, \, \, + \, \, \, \, \, \, \, \, 3 \, \, \, \, \, \, \, \, + ... + n$

$+ S_n = n + (n - 1) + (n - 2) + ... + 1$

So $2S_n = n(n + 1)$.

This is because when you add $S$ to itself (with the order reversed), you get $n + 1$ repeated $n$ times.

Then, $S_n = \displaystyle\frac{n(n + 1)}{2}$

That is $\sum_{k=1}^{n} k = \displaystyle\frac{n(n + 1)}{2}$.

Example $\PageIndex{12}$

Here's that same concept being proven inductively:

Prove that $1 + 2 + ... + n = \displaystyle \frac{n(n + 1)}{2}, \, \forall n \in \mathbb{Z}$

Base step: Choose $n = 1$. Then L.H.S =$1$. and R.H.S $= \frac{(1)(1 + 1)}{2}=1$

Induction Assumption: Assume that $1 + 2 + ... +k= \displaystyle\frac{k(k + 1)}{2}$, for $k \in \mathbb{Z}$.

We shall show that $1 + 2 + ... + k + (k + 1) = \displaystyle\frac{(k + 1)[(k + 1) + 1]}{2} = \frac{(k + 1)(k + 2)}{2}$

Consider $1 + 2 + ... + k + (k + 1)$

$= \displaystyle \frac{k(k + 1)}{2} + (k + 1)$

$= (k + 1) \left( \displaystyle\frac{k}{2} + \displaystyle\frac{1}{1}\right)$

$= (k + 1) \left( \displaystyle\frac{k + 2}{2}\right)$

$= \displaystyle \frac{(k + 1)(k + 2)}{2}$.

Thus, by induction we have $1 + 2 + ... + n = \displaystyle\frac{n(n + 1)}{2}, \, \forall n \in \mathbb{Z}$.

Thinking Out Loud:

Can we determine $S$ of a sequence of integers that do not start at $1$?

The sum of Positive Odd Integers

Thinking Out Loud:

The edges of an equilateral triangle are divided into $n$ equal segments by inserting $n− 1$ points. Lines are drawn through each of these points parallel to each of the three edges, forming a set of small triangles. How many of the small triangles are there? Justify your answer.

Observe:

$1 = 1$

$1 + 3 = 4$

$1 + 3 + 5 = 9$

$1 + 3 + 5 + 7 = 16$

$1 + 3 + 5 + 7 + 9 = 25$

$1 + 3 + 5 + ... + (2n-1) = ?,$

Example $\PageIndex{13}$

Let's look at a table of the sums of the first $n$ positive odd integers:

Table $\PageIndex{13}$: sums of the first $n$ positive odd integers

$n$	$S_n$
1	1
2	4
3	9
4	16
5	25

As we can see, the sum of the first $n$ positive odd integers $= n^2$

Example $\PageIndex{14}$

Let's try a visual proof for this one as well. Remember, square numbers can be arranged into perfectly square arrays.

As we can see, when we arrange odd integers into an array (each new term is represented by a new color), we always have an array with $n^2$ points.

Exercise$\PageIndex{2}$

By using induction, prove that $1 + 3 + 5 + ... + (2n-1) = n^2,$ for all $n \geq1$.

 

The Sum of Positive Even Integers

Observe:

$2 = 2$

$2 + 4 = 6$

$2 + 4 + 6 = 12$

$2 + 4 + 6 + 8 = 20$

$2 + 4 + 6 + 8 + 10 = 30$

$2 + 4 + 6 + ... +2n = ?, n \in \bf{N}$

Example $\PageIndex{15}$

Let's try deriving this using what we already know:

If we write the sum of positive even integers as

$2 + 4 + 6 + 8 + ... + 2n$,

We see we can factor out the 2:

$2(1 + 2 + 3 + ... + n)$. This is great news! We already know the sum of a finite set of positive integers. It is $\displaystyle \frac{n(n + 1)}{2}$

So then the sum of a series of positive even integers is:

$2 \left( \displaystyle \frac{n(n + 1)}{2}\right)$

Or $n (n + 1)$.

Example $\PageIndex{16}$

Let's try a visual proof:

Here is the sum of the first $4$ positive even integers, or $n = 4$:

Now, if we move some of the points to make a rectangular array...

...we can see that, for $n$ terms, our array is described by:

$n(n + 1)$

Contributors

• Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

• Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada)