There are several types of numerical successions, in which for a function we will always have a number as a result for a certain position of that number that we want to calculate. In mathematical terms it is represented like this:

f(n) = x_{n} \quad \text{for} \quad n \ \in \ \mathbb{Q}

\mathbb{Q} = \ \text{rational, integer and natural numbers}

## Finite succession

Finite successions are those that have a starting value and an ending value, for example the following succession:

4, \ 8, \ 12, \ 16, \ 20, \ 24.

It can be seen that the starting value is 4 and the final value is 24.

## Infinite successions

A succession is said to be infinite when it has a starting term but does not have an ending term. Like the following succession:

5, \ 25, \ 125, \ 625, \ \dots

It can be seen that the starting term is 5 and has no ending term due to the ellipsis.

## Genereting Formula

It is necessary that there is a **Generating Formula** or a **Law** so that a succession can be considered as such with which the terms can be generated. The Generating Formula has to be a function of n, with which we can generate any term. Let’s see an example:

f(n) = \cfrac{2n + 1}{2n} \quad \text{for} \quad n = 1, 2, 3, 4, \dots

## Recursive formulas

When a law or formula that generates a succession is given in preceding term or terms and some constant, the formula is said to be recursive.

Recursive formulas relate successive terms in a succession to provide formulas of computing successive quantities in terms of the preceding ones. Next we will see the simple and multiple recursive formulas.

To solve recursive formulas we need to know how to solve systems of equations, we will see later.

### Simple recursive formulas

The Generating Formula is simple recursive when the succession is a function of the previous term and some constant.

t_{k} = \text{A} \ t_{k-1} + \text{B}

Always the t_{1} is the primary term that is needed to generate the next elements of the succession.

### Simple recursive succession example

Given the following succession:

1, 4, 10, 22, 46, \ \dots

To find the generating formula we have to take the general formula of the simple recursive succession:

t_{k} = \text{A}t_{k-1} + \text{B}

And as the primary term we have that it is t_{1}=1, so we have that when k=1, then t_{1}=1.

When we have k=2, we see that t_{2}=\text{A}t_{1}+B, and since t_{2}=4, then 4=\text{A}+\text{B}

When we have k=3, we see that t_{3}=\text{A}t_{2}+\text{B}, and since t_{3}=10, then 10=4\text{A}+\text{B}

So we already have the two systems of equations:

\begin{array}{r} \text{A} + \text{B} = 4 \\ 4\text{A} + \text{B} = 10 \end{array}

When we solve them, we will have the following answers:

\text{A} = 2 \quad \text{y} \quad \text{B} = 2

Finally we have our generating formula:

\begin{array}{ |c| } \hline t_{k} = 2t_{k-1} + 2 \quad \text{with} \quad t_{1} = 1 \quad \text{for} \quad k=2, 3, 4, \dots \\ \hline \end{array}

### Multiple recursive formulas

Generates succession in which the calculated term is a function of two or more previous terms plus some constant, mathematically it is like this:

t_{k} = \text{A} \ t_{k-1} + \text{B} \ t_{k-2} + \text{C}

It can be as complicated as the succession becomes:

t_{k} = \text{A} \ t_{k-1} + \text{B} t_{k-2} + \text{C} t_{k-3} + \text{D}

t_{k} = \text{A} \ t_{k-1} + \text{B} t_{k-2} + \text{C} t_{k-3} + \text{D}t_{k-4} + \text{E}

When we have that the subscript of t is zero or less than zero, then we will take the value that already appears in the succession without considering the constant and the other value of the preceding term.

### Multiple recursive succession example

Determine the generating formula of the following succession:

1, 3, 5, 9, 15, 25, 41, \dots

To begin with, we will suppose that we already did the test and saw that this is not a simple regressive succession and also suppose that the formula is a function of two preceding terms and some constant:

t_{k} = \text{A} t_{k-1} + \text{B} t_{k-2} + \text{C}

Let’s start looking for the generating formula, we have that for k = 1 it is t_ {1} = 1 . Since we have that our t reaches a subscript zero or less than zero, we will then ignore those values for that k = 1 and we will advance to k = 2 , so on until our equations of our system have all their values for the t .

Since we have three preceding terms, then we need three systems to solve a system of three equations.

- For k=3 we have t_{3} = \text{A}t_{2} + \text{B}t_{1} + \text{C}
- For k=4 we have t_{4} = \text{A}t_{3} + \text{B} t_{2} + \text{C}
- For k=5 we have t_{5} = \text{A} t_{4} + \text{B} t_{3} + \text{C}

We have the first values of the succession that go from t_{1} to t_{7}, we are simply substituting them in the previous equations, remember that t_{n} is the position of each value in the given succession:

- For k=3 we have 5 = 3 \text{A} + \text{B}+ \text{C}
- For k=4 we have 9 = 5\text{A} + 3 \text{B} + \text{C}
- For k=5 we have 15 = 9 \text{A} + 5 \text{B} + \text{C}

Now that we have our system of equations, we have to find the values, since we are not going to explain how to solve systems of equations, we go directly with the results:

\text{A} = 1 \qquad \text{B} = 1 \qquad \text{C} = 1

So we already have our generating formula:

\begin{array}{ |c| } \hline t_{k} = t_{k-1} + t_{k-2} + 1 \quad \text{with} \quad t_{1} = 1 \ \text{and} \ t_{2} = 3 \quad \text{for} \quad k=3, 4, 5, \dots \\ \hline \end{array}

#### Note

It may be that we do not know if the succession is simple or multiple recursive, we have to try to obtain our generating formula.

# Geometric succession

It is any number after the first and is obtained by multiplying the previous term by a non-zero number that is called the ratio of the succession:

t_{k} = \text{A} t_{k-1}

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