# Solve the recurrence relation a_n = a_n-1 + 20a_n-2

• hyderman
In summary, the conversation discusses solving a recurrence relation with given initial conditions and finding the final solution using substitution and solving for constants. It also mentions the importance of checking the solution by plugging it into the original recurrence equation.
hyderman
hello
please check my answer and let me know if there are any errors and if there are better ways to solve ...

Solve the recurrence relation

an = an−1 + 20an−2

with a0 = −1 and a1 = 13.

Step – I : Find Characteristic equation and solve for its roots:

an = an−1 + 20an−2 ------equation (0)

First of all make corresponding characteristic equation of equation 0 and find roots of that equation

let
an= r2
an-1 =r
an-2 =r0

Substituting above values in equation (0)
r2 =r + 20

r2-r-20=0 is the characteristic equation.

r2-5r+4r-20=0
r(r-5)+4(r-5)=0
(r-5 ) (r+4)=0

r1 = 5 , r2 = -4 are the roots of this characteristic equation

Step– II : Find Constants’ value using Initial Condition

So solution of this polynomial would be

an = c1 r1n + c2 r2n ------------------equation (1)

Where c1 & c2 are some arbitrary constants .

Now we have to find values of c1 and c2 using Initial Condition.

As a0 = −1 and a1 = 13.

So put n=0 in equation (1)

a0= c1 r10 + c2 r20
a0= c1 + c2
-1 = c1 + c2 ---------equation (2)

Now put n=1 in equation (1)
a1 = c1 r11 + c2 r21

13= c1 (5)1 + c2 (-4)1 --------------equation (3)

Multiply equation (2) by 4

-4= 4c1 + 4c2 -------------------------------------------------equation (4)

Add equation (3) and (4) to get the values of c1 & c2.
-4= 4c1 + 4c2
13= 5c1 - 4c2

9 = 9c1
c1=1

put c1=1 in equation (2)

c2=-2

Step-III ..Substitute values of c1 , c2 , r1 and r2 to get the final solution

so substituting these values in equation
an = 1.5n + (-2) 4n

an = 5n – (2).4n is the final solution of this recurrence relation

Last edited:
an= 5n-2(-4)n into the recurrance equation and see if it satisfies it. It's obvious that it has the right values for a0 and a1.

By the way, if you don't want to use [ sub ] and [ /sub ] (without the spaces) to get subscripts, at least use underscores and parentheses: a_n= a_(n-1)+ 20a_(n-2) is not at all the same as a_n= a_n-1+ 20 a_n- 2!
(and r^n makes more sense than "rn".)

.

So the solution of the recurrence relation a_n = a_n-1 + 20a_n-2 with a0 = −1 and a1 = 13 is:

an = 5n – (2).4n

## 1. What is a recurrence relation?

A recurrence relation is a mathematical equation that defines a sequence of numbers, where each term is defined in terms of previous terms in the sequence.

## 2. How do you solve a recurrence relation?

To solve a recurrence relation, you can use various methods such as substitution, iteration, or generating functions. The specific method used depends on the type and complexity of the recurrence relation.

## 3. What is the purpose of solving a recurrence relation?

The purpose of solving a recurrence relation is to find a closed-form expression for the nth term of a sequence, which allows us to easily calculate any term in the sequence without having to go through all the previous terms.

## 4. How do you solve the specific recurrence relation an = an-1 + 20an-2?

To solve this recurrence relation, we first need to find the characteristic equation by substituting an = rn into the equation. This gives us the equation r2 - r - 20 = 0. Solving for r, we get r = 5 or r = -4. The general solution is then given by an = c1(5)n + c2(-4)n, where c1 and c2 are constants determined by the initial conditions.

## 5. Can this recurrence relation be solved using other methods?

Yes, this specific recurrence relation can also be solved using the method of generating functions, where we convert the recurrence relation into a power series and then solve for the coefficients. The solution obtained using this method will be the same as the one obtained using the characteristic equation method.

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