# Recursive Formula

How to find the terms of a sequence in recursive formula: examples and their solutions.

## Example 1

The recursive formula is a way

to express a sequence.

There are two parts:

Initial term(s): *a*_{1} (or *a*_{2} if needed)

Equation with *a*_{n} and *a*_{n + 1}

*a*_{n + 1} = *a*_{n} + 6

So *a*_{2} = *a*_{1} + 6.*a*_{1} = 4

So *a*_{2} = 4 + 6.

4 + 6 = 10

So *a*_{2} = 10.

*a*_{n + 1} = *a*_{n} + 6

So *a*_{3} = *a*_{2} + 6.*a*_{2} = 10

So *a*_{3} = 10 + 6.

10 + 6 = 16

So *a*_{3} = 16.

*a*_{n + 1} = *a*_{n} + 6

So *a*_{4} = *a*_{3} + 6.*a*_{3} = 16

So *a*_{4} = 16 + 6.

16 + 6 = 22

So *a*_{4} = 22.

*a*_{1} = 6*a*_{2} = 10*a*_{3} = 16*a*_{4} = 22

So the first four terms are

6, 10, 16, and 22.

## Example 2

*a*_{n + 1} = *a*_{n} + 3*n*

So *a*_{2} = *a*_{1} + 3⋅1.*a*_{1} = -1

+3⋅1 = +3

So *a*_{2} = -1 + 3.

-1 + 3 = 2

So *a*_{2} = 2.

*a*_{n + 1} = *a*_{n} + 3*n*

So *a*_{3} = *a*_{2} + 3⋅2.*a*_{2} = 2

+3⋅2 = +6

So *a*_{3} = 2 + 6.

2 + 6 = 8

So *a*_{3} = 8.

*a*_{n + 1} = *a*_{n} + 3*n*

So *a*_{4} = *a*_{3} + 3⋅3.*a*_{3} = 8

+3⋅3 = +9

So *a*_{4} = 8 + 9.

8 + 9 = 17

So *a*_{4} = 17.

*a*_{1} = -1*a*_{2} = 2*a*_{3} = 8*a*_{4} = 17

So the first four terms are

-1, 2, 8, and 17.

## Example 3

*a*_{n + 2} = *a*_{n} + *a*_{n + 1} means

to find the next term,

add the last two terms.

This sequence is called the [Fibonacci Sequence].

*a*_{n + 2} = *a*_{n} + *a*_{n + 1}

So *a*_{3} = *a*_{1} + *a*_{2}.*a*_{1} = 1*a*_{2} = 1

So *a*_{3} = 1 + 1.

1 + 1 = 2

So *a*_{3} = 2.

*a*_{n + 2} = *a*_{n} + *a*_{n + 1}

So *a*_{4} = *a*_{2} + *a*_{3}.*a*_{2} = 1*a*_{3} = 2

So *a*_{4} = 1 + 2.

1 + 2 = 3

So *a*_{4} = 3.

*a*_{n + 2} = *a*_{n} + *a*_{n + 1}

So *a*_{5} = *a*_{3} + *a*_{4}.*a*_{3} = 2*a*_{4} = 3

So *a*_{5} = 2 + 3.

2 + 3 = 5

So *a*_{5} = 5.

*a*_{n + 2} = *a*_{n} + *a*_{n + 1}

So *a*_{6} = *a*_{4} + *a*_{5}.*a*_{4} = 3*a*_{5} = 5

So *a*_{6} = 3 + 5.

3 + 5 = 8

So *a*_{6} = 8.

*a*_{1} = 1*a*_{2} = 1*a*_{3} = 2*a*_{4} = 3*a*_{5} = 5*a*_{6} = 8

So the first six terms are

1, 1, 2, 3, 5, and 8.

## Example 4: Arithmetic Sequence in Recursive Formula

See the given recursive formula:*a*_{n + 1} = *a*_{n} + 6.

It means

to find the next term,

add the last term and +6.

So this is an arithmetic sequence

in recursive formula:*a*_{n + 1} = *a*_{n} + *d*.

Arithmetic sequence

You can directly use [*a*_{1} = 4] and [*d* = 6]

to write *a*_{n}.

But, let's write *a*_{n}

without directly using the formula.

*a*_{n + 1} = *a*_{n} + 6

So *a*_{2} = *a*_{1} + 6.

By the same way,*a*_{3} = *a*_{2} + 6,*a*_{4} = *a*_{3} + 6,

....

And write*a*_{n} = *a*_{n - 1} + 6.

Add these equations.

The gray terms will be cancelled.

The left side is *a*_{n}.

In the right side,*a*_{1} remains.

And (*n* - 1) of 6 remain.

So the right side is*a*_{1} + (*n* - 1)⋅6.

*a*_{1} = 4

So *a*_{n} = 4 + (*n* - 1)⋅6.

As you can see,

this is the *a*_{n} formula of the arithmetic sequence.

(*a*_{1} = 4, *d* = 6)

(*n* - 1)⋅6 = 6*n* - 6

4 - 6 = -2

So *a*_{n} = 6*n* - 2.

## Example 5: Geometric Sequence in Recursive Formula

See the given recursive formula:*a*_{n + 1} = 2*a*_{n}.

It means

to find the next term,

multiply 2 and the last term.

So this is a geometric sequence

in recursive formula:*a*_{n + 1} = *r*⋅*a*_{n}.

Geometric sequence

You can directly use [*a*_{1} = 3] and [*r* = 2]

to write *a*_{n}.

But, let's write *a*_{n}

without directly using the formula.

*a*_{n + 1} = 2*a*_{n}

So *a*_{2} = 2⋅*a*_{1}.

By the same way,*a*_{3} = 2⋅*a*_{2},*a*_{4} = 2⋅*a*_{3},

....

And write*a*_{n} = 2⋅*a*_{n - 1}.

Multiply these equations.

The gray terms will be cancelled.

The left side is *a*_{n}.

In the right side,

(*n* - 1) of 2 remain.

and *a*_{1} remains.

So the right side is

2^{n - 1}⋅*a*_{1}.

*a*_{1} = 3

So *a*_{n} = 2^{n - 1}⋅3.

Switch the order of 2^{n - 1} and 3.

Then *a*_{n} = 3⋅2^{n - 1}.

As you can see,

this is the *a*_{n} formula of the geometric sequence.

(*a*_{1} = 3, *r* = 2)