# 3.1: ArithmeticSequences, Geometric Sequences, Visual Reasoning, and Proof by Induction

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#### 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 \(\PageIndex{1}\):**

\(n\) | \(n^{th}\) 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:**

#### 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 \(\PageIndex{1}\):

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 \(\PageIndex{3}\):**

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 \(\PageIndex{4}\):**

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 \(\PageIndex{5}\):**

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

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

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

#### 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 \(\PageIndex{6}\):**

\(n\) | \(t_n\) | \(t_n / t\)_{\(n - 1\)} |

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 \(\PageIndex{7}\):**

Given the sequence \( -3, 6, -12, 24, -48...\), can we:

- Determine \(a\) and \(r\)
- 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 for finding a general case for the sum of a geometric sequence:

**Example \(\PageIndex{8}\):**

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

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

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

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

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

#### 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 \(\PageIndex{9}\):**

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

**Example \(\PageIndex{10}\):**

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:**

#### Sum of Positive Odd Integers

**Thinking Out Loud:**

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 \(\PageIndex{11}\):**

Let's look at a table of the sums of the first \(n\) postive 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 \(\PageIndex{12}\):**

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 \(\PageIndex{1}\)

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

#### 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 \(\PageIndex{13}\):**

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 \(\PageIndex{14}\):**

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:

\(n(n + 1)\)