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The solution for recurrence relation an =-3an-1 with initial condition a0=2 is

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Answer to Question #144861 in Discrete Mathematics for Ankita

Answers>Math>Discrete Mathematics

Question #144861

Solve the following recurrence relation

a) an = 3an-1 + 4an-2 n≥2 a0=a1=1

b) an= an-2 n≥2 a0=a1=1

c) an= 2an-1 - an-2. n≥2 a0=a1=2

d) an=3an-1 - 3an-2 n≥3 a0=a1=1 , a2=2

Expert's answer

a) a_n = 3a_{n-1} +4a_{n-2}\space n \ge 2, a_0=a_1=1\\a)a

n

=3a

n−1

+4a

n−2

n≥2,a

0

=a

1

=1

Rewrite the recurrence relation a_n - 3a_{n-1} -4a_{n-2} = 0a

n

−3a

n−1

−4a

n−2

=0.

Now form the characteristic equation:

x^2 -3x-4 =0\\ x = -1\space and\space x = 4x

2

−3x−4=0

x=−1 and x=4

We therefore know that the solution to the recurrence

relation will have the form:

a_n = a *(-1)^n +b*4^na

n

=a∗(−1)

n

+b∗4

n

To find a and b , plug in n = 0 and n = 1 to get a system of two equations with two unknowns:

1 = a*(-1)^0+b*4^0 = a+b\\ 1 = a*(-1)^1+b*4^1 =4b - a\\ a = \dfrac{3}{5} \space and \space b = \dfrac{2}{5}\\ \space\\ answer: a_n = \dfrac{3}{5}(-1)^n + \dfrac{2}{5}4^n1=a∗(−1)

0

+b∗4

0

=a+b

1=a∗(−1)

1

+b∗4

1

=4b−a

a=

5

3

and b=

5

2

answer:a

n

=

5

3

(−1)

n

+

5

2

4

n

b) a_n = a_{n-2}\space n \ge 2, a_0=a_1=1\\b)a

n

=a

n−2

n≥2,a

0

=a

1

=1

Rewrite the recurrence relation a_n -a_{n-2} = 0a

n

−a

n−2

=0.

Now form the characteristic equation:

x^2 -1=0\\ x = -1\space and\space x = 1x

2

−1=0

x=−1 and x=1

We therefore know that the solution to the recurrence

relation will have the form:

a_n = a *(-1)^n +b*1^na

n

=a∗(−1)

n

+b∗1

n

To find a and b , plug in n = 0 and n = 1 to get a system of two equations with two unknowns:

1 = a*(-1)^0+b*1^0 = a+b\\ 1 = a*(-1)^1+b*1^1 =b - a\\ a = 0 \space and \space b = 1\\ \space\\ answer: a_n = 1*1^n = 11=a∗(−1)

0

+b∗1

0

=a+b

1=a∗(−1)

1

+b∗1

1

=b−a

a=0 and b=1

answer:a

n

=1∗1

n

=1

c) a_n = 2a_{n-1} -a_{n-2} \space n \ge 2, a_0=a_1=2\\c)a

n

=2a

n−1

−a

n−2

n≥2,a

0

=a

1

=2

Rewrite the recurrence relation a_n - 2a_{n-1} +a_{n-2} = 0a

n

−2a

n−1

+a

n−2

=0.

Now form the characteristic equation:

x^2 -2x+1 =0\\ x = 1x

2

−2x+1=0

x=1

We therefore know that the solution to the recurrence

relation will have the form:

a_n = a *(1)^n +b*n*1^n = a+bna

n

=a∗(1)

n

+b∗n∗1

n

=a+bn

To find a and b , plug in n = 0 and n = 1 to get a system of two equations with two unknowns:

2 = a+b*0 = a\\ 2 = a+b*1 =a +b\\ a = 2 \space and \space b = 0\\ \space\\ answer: a_n = 2 +0*n = 22=a+b∗0=a

2=a+b∗1=a+b

a=2 and b=0

answer:a

n

=2+0∗n=2

d) a_n = 3a_{n-1} -3a_{n-2} \space n \ge 3, a_0=a_1=1, a_2 = 2\\d)a

n

=3a

n−1

−3a

n−2

n≥3,a

0

=a

1

=1,a

2

=2

Rewrite the recurrence relation a_n - 3a_{n-1} +3a_{n-2} = 0a

n

−3a

n−1

+3a

n−2

=0.

Now form the characteristic equation:

x^2 -3x+3 =0\\ x = 1/2(3-i\sqrt3)\ or\\ x = 1/2(3+i\sqrt3)x

2

−3x+3=0

x=1/2(3−i

3

) or

x=1/2(3+i

3

)

We therefore know that the solution to the recurrence

relation will have the form:

a_n = a *(1/2(3-i\sqrt3))^n +b*(1/2(3+i\sqrt3))^na

n

=a∗(1/2(3−i

3

))

n

+b∗(1/2(3+i

3

))

n

To find a and b , plug in n = 0 and n = 1 to get a system of two equations with two unknowns:

1 = a+b \\ 1 = a*(1/2(3-i\sqrt3))+b*(1/2(3+i\sqrt3))\\ 1/2(3+i\sqrt3)-1 = -ai\sqrt3\\ a= 1/6(-3+i\sqrt3)\\ b= 1/6(9-i\sqrt3)\\ \space\\ answer: \\ a_n = 1/6(-3+i\sqrt3) *(1/2(3-i\sqrt3))^n +1/6(9-i\sqrt3)*(1/2(3+i\sqrt3))^n1=a+b

1=a∗(1/2(3−i

3

))+b∗(1/2(3+i

3

))

1/2(3+i

3

)−1=−ai

3

a=1/6(−3+i

3

)

b=1/6(9−i

3

)

answer:

a

n

=1/6(−3+i

3

)∗(1/2(3−i

3

))

n

+1/6(9−i

3

)∗(1/2(3+i

3

))

n

Answered by kumaripayal1850
0

Answer:

Czhmsndvfsmmsnxnxgshjsjsbsbsb

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