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Solved: Find the solution to each of these recurrence
Chapter 1, Problem 17E(choose chapter or problem)
Find the solution to each of these recurrence relations and initial conditions. Use an iterative approach such as that used in Example 10.
Find the solution to each of these recurrence relations and initial conditions. Use an iterative approach such as that used in Example 10.
a) \(a_{n}=3 a_{n-1}, a_{0}=2\)
b) \(a_{n}=a_{n-1}+2, a_{0}=3\)
c) \(a_{n}=a_{n-1}+n, a_{0}=1\)
d) \(a_{n}=a_{n-1}+2 n+3, a_{0}=4\)
e) \(a_{n}=2 a_{n-1}-1, a_{0}=1\)
f) \(a_{n}=3 a_{n-1}+1, a_{0}=1\)
g) \(a_{n}=n a_{n-1}, a_{0}=5\)
h) \(a_{n}=2 n a_{n-1}, a_{0}=1\)
Equation Transcription:
Text Transcription:
a_n = 3a_n-1, a_0=2
a_n =a_n-1 + 2, a_0=3
a_n =a_n-1 + n, a_0=1
a_n =a_n-1 + 2n + 3, a_0=4
a_n = 2a_n-1 - 1, a_0=1
a_n = 3a_n-1 + 1, a_0=1
a_n = na_n-1, a_0=5
a_n = 2na_n-1, a_0=1
Questions & Answers
QUESTION:
Find the solution to each of these recurrence relations and initial conditions. Use an iterative approach such as that used in Example 10.
Find the solution to each of these recurrence relations and initial conditions. Use an iterative approach such as that used in Example 10.
a) \(a_{n}=3 a_{n-1}, a_{0}=2\)
b) \(a_{n}=a_{n-1}+2, a_{0}=3\)
c) \(a_{n}=a_{n-1}+n, a_{0}=1\)
d) \(a_{n}=a_{n-1}+2 n+3, a_{0}=4\)
e) \(a_{n}=2 a_{n-1}-1, a_{0}=1\)
f) \(a_{n}=3 a_{n-1}+1, a_{0}=1\)
g) \(a_{n}=n a_{n-1}, a_{0}=5\)
h) \(a_{n}=2 n a_{n-1}, a_{0}=1\)
Equation Transcription:
Text Transcription:
a_n = 3a_n-1, a_0=2
a_n =a_n-1 + 2, a_0=3
a_n =a_n-1 + n, a_0=1
a_n =a_n-1 + 2n + 3, a_0=4
a_n = 2a_n-1 - 1, a_0=1
a_n = 3a_n-1 + 1, a_0=1
a_n = na_n-1, a_0=5
a_n = 2na_n-1, a_0=1
ANSWER:
Solution :
Step 1:
In this problem we have to find the solution for these recurrence relation, where we given the initial condition.
a: the recurrence relation is given as
an = 3 an-1 with a0= 2
Now we starting with initial condition a0 = 2 and working upward until we reach an to deduce a closed formula for the sequence. Then
We put n = 0 ,1,2,3,4……...so on , Then
a1= 3a1-1
a1 = 3 a0 = 3.2
a2 = 3a2-1
a2 = 3a1 = 3.3.2 = 32.2
a3 = 3a3-1
a3 = 3a2 = 3.32.2 = 33.2
………..
………..
an = 3an-1 = 3n.2