solution of linear homogeneous recurrence relations

Solution of Linear Homogeneous RecurrenceRelations

General Solutions for Homogeneous Problems

Ioan Despi

[email protected]

University of New England

September 16, 2013

Outline

1 Introduction

2 Main TheoremExamples

3 General Procedure

4 More Examples

Ioan Despi – AMTH140 2 of 15

Introduction

If the characteristic equation associated with a given 𝑚-th order linear,constant coefficient, homogeneous recurrence relation has some repeatedroots, then the solution given by

∑︀𝐴𝑖𝜆

𝑛𝑖 will not have 𝑚 arbitrary

constants.

To see this, we assume for instance 𝜆1 = 𝜆2, i.e. root 𝜆1 is repeated.

Then the solution 𝑎𝑛 =𝑚∑︁𝑖=1

𝐴𝑖𝜆𝑛𝑖 = (𝐴1 + 𝐴2)𝜆𝑛

2 + 𝐴3𝜆𝑛3 + · · · + 𝐴𝑚𝜆𝑛

has less than 𝑚 arbitrary constants because 𝐴1 + 𝐴2 comprisesessentially only one arbitrary constant.

To make up for the missing ones, we introduce the following more generaltheorem.

Introduction

∑︀𝐴𝑖𝜆

constants.

Then the solution 𝑎𝑛 =𝑚∑︁𝑖=1

2 + 𝐴3𝜆𝑛3 + · · · + 𝐴𝑚𝜆𝑛

Introduction

∑︀𝐴𝑖𝜆

constants.

Then the solution 𝑎𝑛 =

𝑚∑︁𝑖=1

2 + 𝐴3𝜆𝑛3 + · · · + 𝐴𝑚𝜆𝑛

Introduction

∑︀𝐴𝑖𝜆

constants.

Then the solution 𝑎𝑛 =

𝑚∑︁𝑖=1

2 + 𝐴3𝜆𝑛3 + · · · + 𝐴𝑚𝜆𝑛

Theorem

Suppose the characteristic equation of the linear, constant coefficientrecurrence relation

𝑐𝑚𝑎𝑛+𝑚 + 𝑐𝑚−1𝑎𝑛+𝑚−1 + · · · + 𝑐1𝑎𝑛+1 + 𝑐0𝑎𝑛 = 0, 𝑐𝑚𝑐0 ̸= 0, 𝑛 ≥ 0

has the following roots (all roots accounted for)

𝑚1⏞ ⏟ 𝜆1, · · · , 𝜆1 ,

𝑚2⏞ ⏟ 𝜆2, · · · , 𝜆2 , · · · ,

𝑚𝑘⏞ ⏟ 𝜆𝑘, · · · , 𝜆𝑘

such that 𝜆1, · · · , 𝜆𝑘 are distinct, 𝑚1, · · · ,𝑚𝑘 ≥ 1 and 𝑚1 + · · · + 𝑚𝑘 = 𝑚.Then the general solution of the recurrence relation is

𝑎𝑛 =

𝑘∑︁𝑖=1

⎡⎣(︂𝑚𝑖−1∑︁𝑗=0

𝐴𝑖,𝑗𝑛𝑗

)︂𝜆𝑛𝑖

⎤⎦ , 𝑖.𝑒.,

𝑎𝑛 =(︀𝐴1,0 + 𝐴1,1𝑛 + · · · + 𝐴1,𝑚1−1𝑛

𝑚1−1)︀𝜆𝑛1

+(︀𝐴2,0 + 𝐴2,1𝑛 + · · · + 𝐴2,𝑚2−1𝑛

𝑚2−1))︀𝜆𝑛2 + · · ·

+(︀𝐴𝑘,0 + 𝐴𝑘,1𝑛 + · · · + 𝐴𝑘,𝑚𝑘−1𝑛

𝑚𝑘−1)︀𝜆𝑛𝑘

where 𝐴𝑖,𝑗, for 𝑖 = 1, · · · , 𝑘 and 𝑗 = 0, · · · ,𝑚𝑖 − 1 are arbitrary constants.

So, if the characteristic equation has a root of, say, 𝜆 = 3 that occurstwice (called having multiplicity 2) then in the solution 3𝑛 is multipliedby a polynomial of degree one less that the multiplicity.

I that is of degree one, a linear polynomial: (𝐴𝑛+𝐵)3𝑛.

If the root was 𝜆 = 4 of multiplicity 3 then 4𝑛 is multiplied by apolynomial of degree 2: (𝐴𝑛2 + 𝐵𝑛 + 𝐶)4𝑛.

If the root was 𝜆 = 5 of multiplicity 4 then 5𝑛 is multiplied by apolynomial of degree 3: (𝐴𝑛3 + 𝐵𝑛2 + 𝐶𝑛 + 𝐷)5𝑛.

Example 1

Example

Find a particular solution of

𝑓(𝑛 + 2) + 4𝑓(𝑛 + 1) + 4𝑓(𝑛) = 0, 𝑛 ≥ 0

with initial conditions 𝑓(0) = 1 and 𝑓(1) = 2.

Solution. For clarity, we split the solution procedure into three steps below.

(a) The associated characteristic equation is 𝜆2 + 4𝜆 + 4 = 0 which hasa repeated root 𝜆1 = −2, i.e., 𝑚1 = 2, 𝑘 = 1.

(b) The general solution from the theorem in this lecture is thus

𝑓(𝑛) = (𝐴1,0 + 𝐴1,1𝑛)𝜆𝑛1 = (𝐵0 + 𝐵1𝑛)(−2)𝑛 ,

where 𝐵0 = 𝐴1,0 and 𝐵1 = 𝐴1,1 are arbitrary constants.(c) Constants 𝐵0 and 𝐵1 are to be determined from the initial conditions

𝑓(0) = (𝐵0 + 𝐵1 × 0)(−2)0 = 1𝑓(1) = (𝐵0 + 𝐵1 × 1)(−2)1 = 2

⇔ 𝐵0 = 1−2(𝐵0 + 𝐵1) = 2 .

They are thus 𝐵0 = 1 and 𝐵1 = −2. Hence the requested particular

solution is 𝑓(𝑛) = (1 − 2𝑛)(−2)𝑛 , 𝑛 ≥ 0 .

Example 1

Example

𝑓(𝑛 + 2) + 4𝑓(𝑛 + 1) + 4𝑓(𝑛) = 0, 𝑛 ≥ 0

𝑓(𝑛) = (𝐴1,0 + 𝐴1,1𝑛)𝜆𝑛1 = (𝐵0 + 𝐵1𝑛)(−2)𝑛 ,

𝑓(0) = (𝐵0 + 𝐵1 × 0)(−2)0 = 1𝑓(1) = (𝐵0 + 𝐵1 × 1)(−2)1 = 2

⇔ 𝐵0 = 1−2(𝐵0 + 𝐵1) = 2 .

Example 1

Example

𝑓(𝑛 + 2) + 4𝑓(𝑛 + 1) + 4𝑓(𝑛) = 0, 𝑛 ≥ 0

𝑓(𝑛) = (𝐴1,0 + 𝐴1,1𝑛)𝜆𝑛1 = (𝐵0 + 𝐵1𝑛)(−2)𝑛 ,

where 𝐵0 = 𝐴1,0 and 𝐵1 = 𝐴1,1 are arbitrary constants.

(c) Constants 𝐵0 and 𝐵1 are to be determined from the initial conditions

𝑓(0) = (𝐵0 + 𝐵1 × 0)(−2)0 = 1𝑓(1) = (𝐵0 + 𝐵1 × 1)(−2)1 = 2

⇔ 𝐵0 = 1−2(𝐵0 + 𝐵1) = 2 .

Example 1

Example

𝑓(𝑛 + 2) + 4𝑓(𝑛 + 1) + 4𝑓(𝑛) = 0, 𝑛 ≥ 0

𝑓(𝑛) = (𝐴1,0 + 𝐴1,1𝑛)𝜆𝑛1 = (𝐵0 + 𝐵1𝑛)(−2)𝑛 ,

𝑓(0) = (𝐵0 + 𝐵1 × 0)(−2)0 = 1𝑓(1) = (𝐵0 + 𝐵1 × 1)(−2)1 = 2

⇔ 𝐵0 = 1−2(𝐵0 + 𝐵1) = 2 .

solution is 𝑓(𝑛) = (1 − 2𝑛)(−2)𝑛 , 𝑛 ≥ 0 .Ioan Despi – AMTH140 6 of 15

Example 2

Example

Find the general solution of

𝑎𝑛+3 − 3𝑎𝑛+2 + 4𝑎𝑛 = 0 , 𝑛 ≥ 0

Solution.

The associated characteristic equation is 𝐹 (𝜆)def= 𝜆3 − 3𝜆2 + 4 = 0 .

Although there are formulae to give explicit roots for a third orderpolynomial in terms of its coefficients, we are taking a shortcut here byguessing one root 𝜆1 = −1.

I We can check that 𝜆1 = −1 is indeed a root by verifying𝐹 (−1) = (−1)3 − 3(−1)2 + 4 = 0.

To find the remaining roots, we first factorise 𝐹 (𝜆) by performing a long

division:

Example 2

Example

𝑎𝑛+3 − 3𝑎𝑛+2 + 4𝑎𝑛 = 0 , 𝑛 ≥ 0

Solution.

division:

Example 2

Example

𝑎𝑛+3 − 3𝑎𝑛+2 + 4𝑎𝑛 = 0 , 𝑛 ≥ 0

Solution.

division:

Example 2

Example

𝑎𝑛+3 − 3𝑎𝑛+2 + 4𝑎𝑛 = 0 , 𝑛 ≥ 0

Solution.

division:

Example 2

Example

𝑎𝑛+3 − 3𝑎𝑛+2 + 4𝑎𝑛 = 0 , 𝑛 ≥ 0

Solution.

division:

Example 2

𝜆2 −4𝜆 +4

⎠𝜆 + 1 𝜆3 −3𝜆2 +4

𝜆3 + 𝜆2

−4𝜆2 0−4𝜆2 −4𝜆

4𝜆 +44𝜆 +4

which implies 𝜆3 − 3𝜆2 + 4 = (𝜆 + 1)(𝜆2 − 4𝜆 + 4) = (𝜆 + 1)(𝜆− 2)2.

Therefore all the roots, counting the corresponding multiplicity, are−1 , 2 , 2, i.e. 𝜆1 = −1, 𝑚1 = 1 and 𝜆2 = 2, 𝑚2 = 2.

Hence the general solution reads (with 𝐴 = 𝐴1,0, 𝐵 = 𝐴2,0, 𝐶 = 𝐴2,1)

𝑎𝑛 = 𝐴(−1)𝑛 + (𝐵 + 𝐶𝑛)2𝑛 , 𝑛 ≥ 0 .

Example 2

𝜆2 −4𝜆 +4

⎠𝜆 + 1 𝜆3 −3𝜆2 +4

𝜆3 + 𝜆2

−4𝜆2 0−4𝜆2 −4𝜆

4𝜆 +44𝜆 +4

which implies 𝜆3 − 3𝜆2 + 4 = (𝜆 + 1)(𝜆2 − 4𝜆 + 4) = (𝜆 + 1)(𝜆− 2)2.Therefore all the roots, counting the corresponding multiplicity, are−1 , 2 , 2, i.e. 𝜆1 = −1, 𝑚1 = 1 and 𝜆2 = 2, 𝑚2 = 2.

𝑎𝑛 = 𝐴(−1)𝑛 + (𝐵 + 𝐶𝑛)2𝑛 , 𝑛 ≥ 0 .

Example 2

𝜆2 −4𝜆 +4

⎠𝜆 + 1 𝜆3 −3𝜆2 +4

𝜆3 + 𝜆2

−4𝜆2 0−4𝜆2 −4𝜆

4𝜆 +44𝜆 +4

which implies 𝜆3 − 3𝜆2 + 4 = (𝜆 + 1)(𝜆2 − 4𝜆 + 4) = (𝜆 + 1)(𝜆− 2)2.Therefore all the roots, counting the corresponding multiplicity, are−1 , 2 , 2, i.e. 𝜆1 = −1, 𝑚1 = 1 and 𝜆2 = 2, 𝑚2 = 2.

𝑎𝑛 = 𝐴(−1)𝑛 + (𝐵 + 𝐶𝑛)2𝑛 , 𝑛 ≥ 0 .

Example 3

Example

Find the particular solution of

𝑢𝑛+3 + 3𝑢𝑛+2 + 3𝑢𝑛+1 + 𝑢𝑛 = 0 , 𝑛 ≥ 0

satisfying the initial conditions 𝑢0 = 1, 𝑢1 = 1 and 𝑢2 = −7.

Solution.

(a) The associated characteristic equation 𝜆3 + 3𝜆2 + 3𝜆+ 1 ≡ (𝜆+ 1)3 = 0 has

roots 𝜆1 = −1 with multiplicity 𝑚1 = 3.

(b) The general solution then reads for arbitrary constants 𝐴, 𝐵 and 𝐶

𝑢𝑛 = (𝐴+𝐵𝑛+ 𝐶𝑛2)(−1)𝑛 , 𝑛 ≥ 0

(c) To determine 𝐴,𝐵 and 𝐶 through the use of the initial conditions, we set 𝑛 inthe solution expression in (b) to 0, 1 and 2 respectively, then

𝑢0 gives 𝐴 = 1𝑢1 gives (𝐴+𝐵 + 𝐶)(−1) = 1𝑢2 gives 𝐴+ 2𝐵 + 4𝐶 = −7 .

The solution of these 3 equations, 𝐴 = 1, 𝐵 = 0, 𝐶 = −2. , finally produces

the required particular solution 𝑢𝑛 = (1− 2𝑛2)(−1)𝑛 , 𝑛 ≥ 0

Example 3

Example

𝑢𝑛+3 + 3𝑢𝑛+2 + 3𝑢𝑛+1 + 𝑢𝑛 = 0 , 𝑛 ≥ 0

Solution.

𝑢𝑛 = (𝐴+𝐵𝑛+ 𝐶𝑛2)(−1)𝑛 , 𝑛 ≥ 0

Example 3

Example

𝑢𝑛+3 + 3𝑢𝑛+2 + 3𝑢𝑛+1 + 𝑢𝑛 = 0 , 𝑛 ≥ 0

Solution.

𝑢𝑛 = (𝐴+𝐵𝑛+ 𝐶𝑛2)(−1)𝑛 , 𝑛 ≥ 0

Example 3

Example

𝑢𝑛+3 + 3𝑢𝑛+2 + 3𝑢𝑛+1 + 𝑢𝑛 = 0 , 𝑛 ≥ 0

Solution.

𝑢𝑛 = (𝐴+𝐵𝑛+ 𝐶𝑛2)(−1)𝑛 , 𝑛 ≥ 0

the required particular solution 𝑢𝑛 = (1− 2𝑛2)(−1)𝑛 , 𝑛 ≥ 0Ioan Despi – AMTH140 9 of 15

General Procedure

The general process for solving linear homogeneous recurrence relations withconstant coefficients is like this:

1 Find the characteristic equation.

2 Find the characteristic roots.

3 Find the general solution.

4 Use the initial conditions to find the specific solution.

General Procedure

More Examples

Example

Solve the following (shifted Fibonacci) recurrence relation:

𝑓𝑛 = 𝑓𝑛−1 + 𝑓𝑛−2, 𝑓0 = 0, and 𝑓1 = 1

Solution.

1 The characteristic equation is 𝜆2 − 𝜆− 1 = 0

2 The roots are 𝜆1 = 1+√5

2 and 𝜆2 = 1−√5

3 The general solution is 𝑓𝑛 = 𝐴(︁

1+√5

)︁𝑛

+ 𝐵(︁

1−√5

)︁𝑛

4 Use the initial conditions to find the specific solution

𝑓𝑛 = 1√5

(︁1+

)︁𝑛

− 1√5

(︁1−

)︁𝑛

More Examples

Example

𝑓𝑛 = 𝑓𝑛−1 + 𝑓𝑛−2, 𝑓0 = 0, and 𝑓1 = 1

Solution.

2 and 𝜆2 = 1−√5

1+√5

)︁𝑛

+ 𝐵(︁

1−√5

)︁𝑛

𝑓𝑛 = 1√5

(︁1+

)︁𝑛

− 1√5

(︁1−

)︁𝑛

More Examples

Example

𝑓𝑛 = 𝑓𝑛−1 + 𝑓𝑛−2, 𝑓0 = 0, and 𝑓1 = 1

Solution.

2 and 𝜆2 = 1−√5

1+√5

)︁𝑛

+ 𝐵(︁

1−√5

)︁𝑛

𝑓𝑛 = 1√5

(︁1+

)︁𝑛

− 1√5

(︁1−

)︁𝑛

More Examples

Example

𝑓𝑛 = 𝑓𝑛−1 + 𝑓𝑛−2, 𝑓0 = 0, and 𝑓1 = 1

Solution.

2 and 𝜆2 = 1−√5

1+√5

)︁𝑛

+ 𝐵(︁

1−√5

)︁𝑛

𝑓𝑛 = 1√5

(︁1+

)︁𝑛

− 1√5

(︁1−

)︁𝑛

More Examples

Example

𝑓𝑛 = 𝑓𝑛−1 + 𝑓𝑛−2, 𝑓0 = 0, and 𝑓1 = 1

Solution.

2 and 𝜆2 = 1−√5

1+√5

)︁𝑛

+ 𝐵(︁

1−√5

)︁𝑛

𝑓𝑛 = 1√5

(︁1+

)︁𝑛

− 1√5

(︁1−

)︁𝑛

More Examples

Example

Solve the following recurrence relation:

𝑎𝑛 = 2(𝑎𝑛−1 − 𝑎𝑛−2)

with initial conditions 𝑎0 = 𝑎1 = 1

Solution.

1 The characteristic equation is 𝜆2 − 2𝜆 + 2 = 0

2 The roots are 𝜆1 = 1 + 𝑖 and 𝜆2 = 1 − 𝑖.

3 The general solution is 𝑎𝑛 = 𝐴 (1 + 𝑖)𝑛

+ 𝐵 (1 − 𝑖)𝑛

4 Use the initial conditions to find the specific solution 𝐴 = 1/2, 𝐵 = 1/2,i.e., 𝑎𝑛 = 1

2 (1 + 𝑖)𝑛

+ 12 (1 − 𝑖)

More Examples

Example

𝑎𝑛 = 2(𝑎𝑛−1 − 𝑎𝑛−2)

Solution.

+ 𝐵 (1 − 𝑖)𝑛

2 (1 + 𝑖)𝑛

+ 12 (1 − 𝑖)

More Examples

Example

𝑎𝑛 = 2(𝑎𝑛−1 − 𝑎𝑛−2)

Solution.

+ 𝐵 (1 − 𝑖)𝑛

2 (1 + 𝑖)𝑛

+ 12 (1 − 𝑖)

More Examples

Example

𝑎𝑛 = 2(𝑎𝑛−1 − 𝑎𝑛−2)

Solution.

+ 𝐵 (1 − 𝑖)𝑛

2 (1 + 𝑖)𝑛

+ 12 (1 − 𝑖)

More Examples

Example

𝑎𝑛 = 2(𝑎𝑛−1 − 𝑎𝑛−2)

Solution.

+ 𝐵 (1 − 𝑖)𝑛

2 (1 + 𝑖)𝑛

+ 12 (1 − 𝑖)

More Examples

Example

𝑎𝑛 = 6𝑎𝑛−1 − 9𝑎𝑛−2

with initial conditions 𝑎0 = 3 and 𝑎1 = 10

Solution.

2 The roots for (𝜆− 3)2 = 0 are 𝜆1 = 𝜆2 = 3, so 𝑚1 = 2.

3 The general solution is 𝑎𝑛 = (𝐴 + 𝐵𝑛)3𝑛

4 Use the initial conditions to find the specific solution𝑎𝑛 =

(︀3 + 1

3𝑛)︀

3𝑛 = 3𝑛+1 + 𝑛3𝑛−1

More Examples

Example

𝑎𝑛 = 6𝑎𝑛−1 − 9𝑎𝑛−2

Solution.

(︀3 + 1

3𝑛)︀

3𝑛 = 3𝑛+1 + 𝑛3𝑛−1

More Examples

Example

𝑎𝑛 = 6𝑎𝑛−1 − 9𝑎𝑛−2

Solution.

(︀3 + 1

3𝑛)︀

3𝑛 = 3𝑛+1 + 𝑛3𝑛−1

More Examples

Example

𝑎𝑛 = 6𝑎𝑛−1 − 9𝑎𝑛−2

Solution.

(︀3 + 1

3𝑛)︀

3𝑛 = 3𝑛+1 + 𝑛3𝑛−1

More Examples

Example

𝑎𝑛 = 6𝑎𝑛−1 − 9𝑎𝑛−2

Solution.

(︀3 + 1

3𝑛)︀

3𝑛 = 3𝑛+1 + 𝑛3𝑛−1

More Examples

Example

2𝑎𝑛+3 = 𝑎𝑛+2 + 2𝑎𝑛+1 − 𝑎𝑛

with initial conditions 𝑎0 = 0, 𝑎1 = 1 and 𝑎2 = 2

Solution.

1 The characteristic equation is 2𝜆3 − 𝜆2 − 2𝜆 + 1 = 0

2 The roots are 𝜆1 = 1, 𝜆2 = −1, 𝜆3 = 1/2

3 The general solution is 𝑎𝑛 = 𝐴(1)𝑛 + 𝐵(−1)𝑛 + 𝐶( 12 )𝑛

4 Use the initial conditions to find the specific solution 𝐴 = 5/2, 𝐵 = 1/6,𝐶 = −8/3, so 𝑎𝑛 = 5

2 + 16 (−1)𝑛 − 8

3 ( 12 )𝑛

More Examples

Example

2𝑎𝑛+3 = 𝑎𝑛+2 + 2𝑎𝑛+1 − 𝑎𝑛

Solution.

2 The roots are 𝜆1 = 1, 𝜆2 = −1, 𝜆3 = 1/2

2 + 16 (−1)𝑛 − 8

3 ( 12 )𝑛

More Examples

Example

2𝑎𝑛+3 = 𝑎𝑛+2 + 2𝑎𝑛+1 − 𝑎𝑛

Solution.

2 The roots are 𝜆1 = 1, 𝜆2 = −1, 𝜆3 = 1/2

2 + 16 (−1)𝑛 − 8

3 ( 12 )𝑛

More Examples

Example

2𝑎𝑛+3 = 𝑎𝑛+2 + 2𝑎𝑛+1 − 𝑎𝑛

Solution.

2 The roots are 𝜆1 = 1, 𝜆2 = −1, 𝜆3 = 1/2

2 + 16 (−1)𝑛 − 8

3 ( 12 )𝑛

More Examples

Example

2𝑎𝑛+3 = 𝑎𝑛+2 + 2𝑎𝑛+1 − 𝑎𝑛

Solution.

2 The roots are 𝜆1 = 1, 𝜆2 = −1, 𝜆3 = 1/2

2 + 16 (−1)𝑛 − 8

3 ( 12 )𝑛

More Examples

Example

How many binary strings (strings of ’0’s and ’1’s) of 𝑛 bits have no twoconsecutive zeros?

Solution.

Let 𝑎𝑛 be the number of such strings.

Clearly, 𝑎0 = 0, 𝑎1 = 2, and 𝑎2 = 3.

An 𝑛 bit such string

I begins with (1 , and ends with any 𝑛− 1 bit string, or

I begins with (0, 1 , and ends with any 𝑛− 2 bit string.

So 𝑎𝑛 = 𝑎𝑛−1 + 𝑎𝑛−2 .

The characteristic equation is 𝜆2 − 𝜆− 1, etc.

More Examples

Example

Solution.

Clearly, 𝑎0 = 0, 𝑎1 = 2, and 𝑎2 = 3.

So 𝑎𝑛 = 𝑎𝑛−1 + 𝑎𝑛−2 .

More Examples

Example

Solution.

Clearly, 𝑎0 = 0, 𝑎1 = 2, and 𝑎2 = 3.

So 𝑎𝑛 = 𝑎𝑛−1 + 𝑎𝑛−2 .

More Examples

Example

Solution.

Clearly, 𝑎0 = 0, 𝑎1 = 2, and 𝑎2 = 3.

So 𝑎𝑛 = 𝑎𝑛−1 + 𝑎𝑛−2 .

More Examples

Example

Solution.

Clearly, 𝑎0 = 0, 𝑎1 = 2, and 𝑎2 = 3.

So 𝑎𝑛 = 𝑎𝑛−1 + 𝑎𝑛−2 .

More Examples

Example

Solution.

Clearly, 𝑎0 = 0, 𝑎1 = 2, and 𝑎2 = 3.

So 𝑎𝑛 = 𝑎𝑛−1 + 𝑎𝑛−2 .

More Examples

Example

Solution.

Clearly, 𝑎0 = 0, 𝑎1 = 2, and 𝑎2 = 3.

So 𝑎𝑛 = 𝑎𝑛−1 + 𝑎𝑛−2 .

More Examples

Example

Solution.

Clearly, 𝑎0 = 0, 𝑎1 = 2, and 𝑎2 = 3.

So 𝑎𝑛 = 𝑎𝑛−1 + 𝑎𝑛−2 .

solution of linear homogeneous recurrence relations

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