formula for sequences satisfying second order recurrence relations

formula for sequences satisfying second order recurrence relations

Theorem

.

Let

{

s

n

}

be a

sequence

such that there exist

constants

A

,

B

∈

C

with

s

1

=

A

⁢

s

0

and, for all

positive integers

n

,

s

n

+

1

=

A

⁢

s

n

+

B

⁢

s

n

-

1

.

Then for all

nonnegative integers

n

, we have

s

n

=

s

0

⁢

∑

k

=

0

⌊

n

2

⌋

(

n

-

k

k

)

⁢

B

k

⁢

A

n

-

2

⁢

k

,

where

⌊

⋅

⌋

denotes the

floor function

and

(

a

r

)

denotes the

binomial coefficient

.

Proof.

This will be proven by

induction

. Due to the presence of

⌊

n

2

⌋

, odd

n

and even

n

should be considered separately. Thus, there are two

base steps

:

•

If

n

=

0

, then

s

0

=

s

0

⁢

∑

k

=

0

0

(

0

-

k

k

)

⁢

B

k

⁢

A

0

-

2

⁢

k

.

•

If

n

=

1

, then

s

1

=

A

⁢

s

0

=

s

0

⁢

∑

k

=

0

0

(

1

-

k

k

)

⁢

B

k

⁢

A

1

-

2

⁢

k

.

Now assume that the

theorem

holds for all nonnegative integers less than or equal to some positive integer

n

. We must show that the theorem holds for

n

+

1

.

Recall that

s

n

+

1

=

A

⁢

s

n

+

B

⁢

s

n

-

1

.

We can use the induction hypothesis on

s

n

and

s

n

-

1

:

s

n

+

1

=

A

⁢

s

0

⁢

∑

k

=

0

⌊

n

2

⌋

(

n

-

k

k

)

⁢

B

k

⁢

A

n

-

2

⁢

k

+

B

⁢

s

0

⁢

∑

k

=

0

⌊

n

-

1

2

⌋

(

n

-

1

-

k

k

)

⁢

B

k

⁢

A

n

-

1

-

2

⁢

k

=

s

0

⁢

(

∑

k

=

0

⌊

n

2

⌋

(

n

-

k

k

)

⁢

B

k

⁢

A

n

+

1

-

2

⁢

k

+

∑

k

=

0

⌊

n

-

1

2

⌋

(

n

-

1

-

k

k

)

⁢

B

k

+

1

⁢

A

n

-

1

-

2

⁢

k

)

=

s

0

⁢

(

A

n

+

1

+

∑

k

=

1

⌊

n

2

⌋

(

n

-

k

k

)

⁢

B

k

⁢

A

n

+

1

-

2

⁢

k

+

∑

k

=

1

⌊

n

+

1

2

⌋

(

n

-

k

k

-

1

)

⁢

B

k

⁢

A

n

+

1

-

2

⁢

k

)

First assume that

n

is odd. Then the first of the two

summations

has

⌊

n

2

⌋

=

n

-

1

2

terms and the second summation has

⌊

n

+

1

2

⌋

=

n

+

1

2

terms. Therefore, we must split off the

k

=

n

+

1

2

term from the second summation before combining the two summations:

s

n

+

1

=

s

0

⁢

(

A

n

+

1

+

∑

k

=

1

⌊

n

2

⌋

(

n

-

k

k

)

⁢

B

k

⁢

A

n

+

1

-

2

⁢

k

+

∑

k

=

1

⌊

n

2

⌋

(

n

-

k

k

-

1

)

⁢

B

k

⁢

A

n

+

1

-

2

⁢

k

+

B

n

+

1

2

)

=

s

0

⁢

(

A

n

+

1

+

∑

k

=

1

⌊

n

2

⌋

[

(

n

-

k

k

)

+

(

n

-

k

k

-

1

)

]

⁢

B

k

⁢

A

n

+

1

-

2

⁢

k

+

B

n

+

1

2

)

=

s

0

⁢

(

A

n

+

1

+

∑

k

=

1

⌊

n

2

⌋

(

n

+

1

-

k

k

)

⁢

B

k

⁢

A

n

+

1

-

2

⁢

k

+

B

n

+

1

2

)

=

s

0

⁢

∑

k

=

0

⌊

n

+

1

2

⌋

(

n

+

1

-

k

k

)

⁢

B

k

⁢

A

n

+

1

-

2

⁢

k

Now assume that

n

is even. Then the two summations have the same number of terms and can be combined as is:

s

n

+

1

=

s

0

⁢

(

A

n

+

1

+

∑

k

=

1

⌊

n

+

1

2

⌋

[

(

n

-

k

k

)

+

(

n

-

k

k

-

1

)

]

⁢

B

k

⁢

A

n

+

1

-

2

⁢

k

)

=

s

0

⁢

∑

k

=

0

⌊

n

+

1

2

⌋

(

n

+

1

-

k

k

)

⁢

B

k

⁢

A

n

+

1

-

2

⁢

k

Hence, the theorem holds for all nonnegative integers

n

.
∎

Title

formula for sequences satisfying second order recurrence relations

Canonical

name

FormulaForSequencesSatisfyingSecondOrderRecurrenceRelations

Date of creation

2013-03-22 17:51:43

Last modified on

2013-03-22 17:51:43

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

7

Author

Wkbj79 (1863)

Entry type

Theorem

Classification

msc 11B37

Classification

msc 03D20

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