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hyderman

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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

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

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