# Recursive Formula

How to solve recursive formula problems: 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)

Difference equation (*a*_{n} and *a*_{n + 1})

To find *a*_{2},

put *a*_{1} = 4

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

= 10

To find *a*_{3},

put *a*_{2} = 10

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

= 16

To find *a*_{4},

put *a*_{3} = 16

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

= 22

So the first four terms are 4, 10, 16, and 22.

## Example 2

To find *a*_{2},

put *a*_{1} = 3

into *a*_{2} = 2*a*_{1}.*a*_{2} = 2⋅3

= 6

To find *a*_{3},

put *a*_{2} = 6

into *a*_{3} = 2*a*_{2}.*a*_{3} = 2⋅6

= 12

To find *a*_{4},

put *a*_{3} = 12

into *a*_{4} = 2*a*_{3}.*a*_{4} = 2⋅12

= 24

So the first four terms are 3, 6, 12, and 24.

## Example 3

To find *a*_{2},

put *a*_{1} = -1

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

= 2

To find *a*_{3},

put *a*_{2} = 2

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

= 8

To find *a*_{4},

put *a*_{3} = 8

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

= 17

So the first four terms are -1, 2, 8, and 17.

## Example 4

To find *a*_{3},

put *a*_{1} = 1 and *a*_{1} = 1

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

= 2

To find *a*_{4},

put *a*_{2} = 1 and *a*_{3} = 2

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

= 3

To find *a*_{5},

put *a*_{3} = 2 and *a*_{4} = 3

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

= 5

To find *a*_{6},

put *a*_{4} = 3 and *a*_{5} = 5

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

= 8

So the first six terms are 1, 1, 2, 3, 5, and 8.

## Example 5

The difference between *a*_{n + 1} and *a*_{n} is constant: 6.

So this recursive formula

shows an arithmetic sequence.

Write the difference equations vertically,

starting from *a*_{2} to *a*_{n}.*a*_{2} = *a*_{1} + 6*a*_{3} = *a*_{2} + 6*a*_{4} = *a*_{3} + 6

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

Then add these equations.

The gray terms on both sides are cancelled.

(*a*_{2} ~ *a*_{n - 1})

And '+6' is added '*n* - 1' times.

So *a*_{n} = *a*_{1} + (*n* - 1)⋅6.

Arrange the terms.

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

## Example 6

The ratio between *a*_{n + 1} and *a*_{n} is constant: 2.

So this recursive formula

shows a geometric sequence.

Write the difference equations vertically,

starting from *a*_{2} to *a*_{n}.*a*_{2} = 2*a*_{1}*a*_{3} = 2*a*_{2}*a*_{4} = 2*a*_{3}

...*a*_{n} = 2*a*_{n - 1}

Then multiply these equations.

The gray terms on both sides are cancelled.

(*a*_{2} ~ *a*_{n - 1})

And '2' is multiplied '*n* - 1' times.

So *a*_{n} = *a*_{1}⋅2^{n - 1}.

Arrange the terms.

Then *a*_{n} = 3⋅2^{n - 1}.