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Determine the solution for the recurrence relation bn = 8bn-1 − 12bn-2 with b0 = 3 and b1 = 4.
- a)7/2*2n − 1/2*6n
- b)2/3*7n - 5*4n
- c)4!*6n
- d)2/8n
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
Determine the solution for the recurrence relation bn= 8bn-1− 12...
Rewrite the recurrence relation bn - 8bn-1 + 12bn-2 = 0. Now from the characteristic equation: x2 − 8x + 12 = 0 we have x: (x−2)(x−6) = 0, so x = 2 and x = 6 are the characteristic roots. Therefore the solution to the recurrence relation will have the form: bn = b2n + c6n. To find b and c, set n = 0 and n = 1 to get a system of two equations with two unknowns: 3 = b20 + c60 = b + c, and 4 = b21 + c61 = 2b + 6c. Solving this system gives c=-1/2 and b = 7/2. So the solution to the recurrence relation is, bn = 7/2*2n − 1/2*6n.
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Determine the solution for the recurrence relation bn= 8bn-1− 12...
Understanding the Recurrence Relation
The given recurrence relation is:
bn = 8bn-1 - 12bn-2
with initial conditions b0 = 3 and b1 = 4.
We will solve this by finding the characteristic equation.
Characteristic Equation
To find the characteristic equation, we assume a solution of the form bn = r^n. Substituting this into the recurrence gives:
r^n = 8r^(n-1) - 12r^(n-2)
Dividing through by r^(n-2) (for r ≠ 0):
r^2 = 8r - 12
This simplifies to the characteristic equation:
r^2 - 8r + 12 = 0
Finding Roots
We can factor this quadratic equation:
(r - 6)(r - 2) = 0
Thus, the roots are r1 = 6 and r2 = 2.
General Solution
The general solution for the recurrence relation is:
bn = A * 6^n + B * 2^n
where A and B are constants determined by the initial conditions.
Applying Initial Conditions
Using b0 = 3:
3 = A * 6^0 + B * 2^0
3 = A + B (1)
Using b1 = 4:
4 = A * 6^1 + B * 2^1
4 = 6A + 2B (2)
Now, we can solve these equations:
From (1), B = 3 - A. Substituting into (2):
4 = 6A + 2(3 - A)
4 = 6A + 6 - 2A
4 = 4A + 6
4A = -2
A = -1/2
Substituting A back into (1):
B = 3 - (-1/2) = 3 + 1/2 = 7/2.
Final Solution
Thus, the solution is:
bn = (7/2) * 2^n - (1/2) * 6^n.
This corresponds to option 'A'.
The given recurrence relation is:
bn = 8bn-1 - 12bn-2
with initial conditions b0 = 3 and b1 = 4.
We will solve this by finding the characteristic equation.
Characteristic Equation
To find the characteristic equation, we assume a solution of the form bn = r^n. Substituting this into the recurrence gives:
r^n = 8r^(n-1) - 12r^(n-2)
Dividing through by r^(n-2) (for r ≠ 0):
r^2 = 8r - 12
This simplifies to the characteristic equation:
r^2 - 8r + 12 = 0
Finding Roots
We can factor this quadratic equation:
(r - 6)(r - 2) = 0
Thus, the roots are r1 = 6 and r2 = 2.
General Solution
The general solution for the recurrence relation is:
bn = A * 6^n + B * 2^n
where A and B are constants determined by the initial conditions.
Applying Initial Conditions
Using b0 = 3:
3 = A * 6^0 + B * 2^0
3 = A + B (1)
Using b1 = 4:
4 = A * 6^1 + B * 2^1
4 = 6A + 2B (2)
Now, we can solve these equations:
From (1), B = 3 - A. Substituting into (2):
4 = 6A + 2(3 - A)
4 = 6A + 6 - 2A
4 = 4A + 6
4A = -2
A = -1/2
Substituting A back into (1):
B = 3 - (-1/2) = 3 + 1/2 = 7/2.
Final Solution
Thus, the solution is:
bn = (7/2) * 2^n - (1/2) * 6^n.
This corresponds to option 'A'.
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