The solution for recurrence relation an =-3an-1 with initial condition a0=2 is
Answers
Step-by-step explanation:
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
Answer:
Step-by-step explanation:
To solve the recurrence relation an = 3an-1 with the initial condition a0 = 2, we can use a recursive approach.
Given:
a0 = 2
an = 3an-1
To find a1, we substitute n = 1 into the recurrence relation:
a1 = 3a1-1 = 3a0 = 3(2) = 6
To find a2, we substitute n = 2 into the recurrence relation:
a2 = 3a2-1 = 3a1 = 3(6) = 18
To find a3, we substitute n = 3 into the recurrence relation:
a3 = 3a3-1 = 3a2 = 3(18) = 54
We can observe a pattern in the recursive relation:
a1 = 3a0 = 3(2) = 6
a2 = 3a1 = 3(6) = 18
a3 = 3a2 = 3(18) = 54
From the pattern, it can be observed that the nth term an is given by:
an = 3^n * a0
Substituting a0 = 2, we have:
an = 3^n * 2
Therefore, the solution to the recurrence relation an = 3an-1 with the initial condition a0 = 2 is given by an = 3^n * 2.