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Determine the solution of the recurrence relation Fn = 20Fn-1 − 25Fn-2 where F0 = 4 and F1 = 14.
- a)an = 14*5n-1
- b)an = 7/2*2n−1/2*6n
- c)an = 7/2*2n−3/4*6n+1
- d)an = 3*2n−1/2*3n
Correct answer is option 'B'. Can you explain this answer?
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Determine the solution of the recurrence relation Fn= 20Fn-1− 25...
The characteristic equation of the recurrence relation is → x2−20x + 36 = 0
So, (x-2)(x-18) = 0. Hence, there are two real roots x1=2 and x2=18. Therefore the solution to the recurrence relation will have the form: an = a2n+b18n. To find a and b, set n = 0 and n = 1 to get a system of two equations with two unknowns: 4 = a20 + b180 = a + b and 3=a21 + b61 = 2a + 6b. Solving this system gives b = -1/2 and a = 7/2. So the solution to the recurrence relation is,
an = 7/2*2n−1/2*6n.
So, (x-2)(x-18) = 0. Hence, there are two real roots x1=2 and x2=18. Therefore the solution to the recurrence relation will have the form: an = a2n+b18n. To find a and b, set n = 0 and n = 1 to get a system of two equations with two unknowns: 4 = a20 + b180 = a + b and 3=a21 + b61 = 2a + 6b. Solving this system gives b = -1/2 and a = 7/2. So the solution to the recurrence relation is,
an = 7/2*2n−1/2*6n.
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Determine the solution of the recurrence relation Fn= 20Fn-1− 25...
To find the solution of the recurrence relation Fn = 20Fn-1, we can start by finding the base case, which is usually given or can be derived from the problem statement.
Let's assume that the base case is F0 = 1 (this can be changed depending on the problem).
Now, let's plug in some values to see the pattern:
F1 = 20F0 = 20
F2 = 20F1 = 20(20) = 400
F3 = 20F2 = 20(400) = 8000
F4 = 20F3 = 20(8000) = 160000
...
From these calculations, we can see that Fn = 20^n.
Therefore, the solution to the recurrence relation Fn = 20Fn-1 is Fn = 20^n.
Let's assume that the base case is F0 = 1 (this can be changed depending on the problem).
Now, let's plug in some values to see the pattern:
F1 = 20F0 = 20
F2 = 20F1 = 20(20) = 400
F3 = 20F2 = 20(400) = 8000
F4 = 20F3 = 20(8000) = 160000
...
From these calculations, we can see that Fn = 20^n.
Therefore, the solution to the recurrence relation Fn = 20Fn-1 is Fn = 20^n.
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Determine the solution of the recurrence relation Fn= 20Fn-1− 25Fn-2where F0= 4 and F1= 14.a)an= 14*5n-1b)an= 7/2*2n−1/2*6nc)an= 7/2*2n−3/4*6n+1d)an= 3*2n−1/2*3nCorrect answer is option 'B'. Can you explain this answer?
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Determine the solution of the recurrence relation Fn= 20Fn-1− 25Fn-2where F0= 4 and F1= 14.a)an= 14*5n-1b)an= 7/2*2n−1/2*6nc)an= 7/2*2n−3/4*6n+1d)an= 3*2n−1/2*3nCorrect answer is option 'B'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Determine the solution of the recurrence relation Fn= 20Fn-1− 25Fn-2where F0= 4 and F1= 14.a)an= 14*5n-1b)an= 7/2*2n−1/2*6nc)an= 7/2*2n−3/4*6n+1d)an= 3*2n−1/2*3nCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Determine the solution of the recurrence relation Fn= 20Fn-1− 25Fn-2where F0= 4 and F1= 14.a)an= 14*5n-1b)an= 7/2*2n−1/2*6nc)an= 7/2*2n−3/4*6n+1d)an= 3*2n−1/2*3nCorrect answer is option 'B'. Can you explain this answer?.
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ample number of questions to practice Determine the solution of the recurrence relation Fn= 20Fn-1− 25Fn-2where F0= 4 and F1= 14.a)an= 14*5n-1b)an= 7/2*2n−1/2*6nc)an= 7/2*2n−3/4*6n+1d)an= 3*2n−1/2*3nCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice Civil Engineering (CE) tests.
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