Civil Engineering (CE) Exam > Civil Engineering (CE) Questions > If Sn= 4Sn-1+ 12n, where S0= 6 and S1= 7, fin...
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If Sn = 4Sn-1 + 12n, where S0 = 6 and S1 = 7, find the solution for the recurrence relation.
- a)an = 7(2n)−29/6n6n
- b)an = 6(6n) + 6/7n6n
- c)an = 6(3n+1)−5n
- d)an = nn − 2/6n6n
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
If Sn= 4Sn-1+ 12n, where S0= 6 and S1= 7, find the solution for the re...
First, we need to find a pattern in the recurrence relation. Let's start by finding S2:
S2 = 4S1 + 12(2) = 4(7) + 24 = 52
Now let's find S3:
S3 = 4S2 + 12(3) = 4(52) + 36 = 220
We can continue this pattern to find Sn in terms of Sn-1:
Sn = 4Sn-1 + 12n
Substituting Sn-1 into this equation, we get:
Sn = 4(4Sn-2 + 12(n-1)) + 12n
Sn = 16Sn-2 + 48n + 12n
Sn = 16(4Sn-3 + 12(n-2)) + 48n + 12n
Sn = 64Sn-3 + 192n + 48n
Sn = 64(4Sn-4 + 12(n-3)) + 192n + 48n
...
We can see that the pattern continues with each term having a factor of 4 and a constant term of 12n. So, we can write:
Sn = 4kSn-k + 12n + 12n + ... + 12n (k times)
where k is the number of times we need to apply the recurrence relation to get to Sn.
To find k, we can set n = 1 and use the given values:
S1 = 4S0 + 12(1)
7 = 4(6) + 12
k = 1
So, we need to apply the recurrence relation once to get to Sn. Therefore:
Sn = 4S(n-1) + 12n
Sn = 4(4S(n-2) + 12(n-1)) + 12n
Sn = 16S(n-2) + 48n + 12n
Sn = 16(4S(n-3) + 12(n-2)) + 48n + 12n
...
Sn = 4^n S0 + 12(1 + 2 + 3 + ... + n)
Using the formula for the sum of the first n natural numbers, we get:
Sn = 4^n(6) + 6n(n+1)
Therefore, the solution for the recurrence relation is:
an = 4^n(6) + 6n(n+1)
S2 = 4S1 + 12(2) = 4(7) + 24 = 52
Now let's find S3:
S3 = 4S2 + 12(3) = 4(52) + 36 = 220
We can continue this pattern to find Sn in terms of Sn-1:
Sn = 4Sn-1 + 12n
Substituting Sn-1 into this equation, we get:
Sn = 4(4Sn-2 + 12(n-1)) + 12n
Sn = 16Sn-2 + 48n + 12n
Sn = 16(4Sn-3 + 12(n-2)) + 48n + 12n
Sn = 64Sn-3 + 192n + 48n
Sn = 64(4Sn-4 + 12(n-3)) + 192n + 48n
...
We can see that the pattern continues with each term having a factor of 4 and a constant term of 12n. So, we can write:
Sn = 4kSn-k + 12n + 12n + ... + 12n (k times)
where k is the number of times we need to apply the recurrence relation to get to Sn.
To find k, we can set n = 1 and use the given values:
S1 = 4S0 + 12(1)
7 = 4(6) + 12
k = 1
So, we need to apply the recurrence relation once to get to Sn. Therefore:
Sn = 4S(n-1) + 12n
Sn = 4(4S(n-2) + 12(n-1)) + 12n
Sn = 16S(n-2) + 48n + 12n
Sn = 16(4S(n-3) + 12(n-2)) + 48n + 12n
...
Sn = 4^n S0 + 12(1 + 2 + 3 + ... + n)
Using the formula for the sum of the first n natural numbers, we get:
Sn = 4^n(6) + 6n(n+1)
Therefore, the solution for the recurrence relation is:
an = 4^n(6) + 6n(n+1)
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Community Answer
If Sn= 4Sn-1+ 12n, where S0= 6 and S1= 7, find the solution for the re...
The characteristic equation of the recurrence relation is → x2 − 4x - 12 = 0
So, (x-6)(x+2) = 0. Only the characteristic root is 6. Therefore the solution to the recurrence relation will have the form: an = a.6n+b.n.6n. To find a and b, set n=0 and n=1 to get a system of two equations with two unknowns: 6 = a60 + b.0.60 = a and 7 = a61 + b.1.61 = 2a + 6b. Solving this system gives a=6 and b=6/7. So the solution to the recurrence relation is, an = 6(6n) − 6/7n6n.
So, (x-6)(x+2) = 0. Only the characteristic root is 6. Therefore the solution to the recurrence relation will have the form: an = a.6n+b.n.6n. To find a and b, set n=0 and n=1 to get a system of two equations with two unknowns: 6 = a60 + b.0.60 = a and 7 = a61 + b.1.61 = 2a + 6b. Solving this system gives a=6 and b=6/7. So the solution to the recurrence relation is, an = 6(6n) − 6/7n6n.
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If Sn= 4Sn-1+ 12n, where S0= 6 and S1= 7, find the solution for the recurrence relation.a)an= 7(2n)−29/6n6nb)an= 6(6n) + 6/7n6nc)an= 6(3n+1)−5nd)an= nn − 2/6n6nCorrect answer is option 'B'. Can you explain this answer?
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If Sn= 4Sn-1+ 12n, where S0= 6 and S1= 7, find the solution for the recurrence relation.a)an= 7(2n)−29/6n6nb)an= 6(6n) + 6/7n6nc)an= 6(3n+1)−5nd)an= nn − 2/6n6nCorrect 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 If Sn= 4Sn-1+ 12n, where S0= 6 and S1= 7, find the solution for the recurrence relation.a)an= 7(2n)−29/6n6nb)an= 6(6n) + 6/7n6nc)an= 6(3n+1)−5nd)an= nn − 2/6n6nCorrect 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 If Sn= 4Sn-1+ 12n, where S0= 6 and S1= 7, find the solution for the recurrence relation.a)an= 7(2n)−29/6n6nb)an= 6(6n) + 6/7n6nc)an= 6(3n+1)−5nd)an= nn − 2/6n6nCorrect answer is option 'B'. Can you explain this answer?.
If Sn= 4Sn-1+ 12n, where S0= 6 and S1= 7, find the solution for the recurrence relation.a)an= 7(2n)−29/6n6nb)an= 6(6n) + 6/7n6nc)an= 6(3n+1)−5nd)an= nn − 2/6n6nCorrect 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 If Sn= 4Sn-1+ 12n, where S0= 6 and S1= 7, find the solution for the recurrence relation.a)an= 7(2n)−29/6n6nb)an= 6(6n) + 6/7n6nc)an= 6(3n+1)−5nd)an= nn − 2/6n6nCorrect 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 If Sn= 4Sn-1+ 12n, where S0= 6 and S1= 7, find the solution for the recurrence relation.a)an= 7(2n)−29/6n6nb)an= 6(6n) + 6/7n6nc)an= 6(3n+1)−5nd)an= nn − 2/6n6nCorrect answer is option 'B'. Can you explain this answer?.
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