Math Problem Statement
Find the solution to each of these recurrence relations with the given initial conditions. Use an iterative approach. a) a n =-a n-1 , a_{0} = 5 b ) a n =a n-1 +3 , a_{0} = 1 ) a n =a n-1 -n , a_{0} = 4 d) a n =2a n-1 -3 , a_{0} = - 1 e) a n =(n+1)a n-1 , a_{0} = 2 f) an 2nan a_{0} = 3
Solution
Let's solve each of the recurrence relations using an iterative approach and the given initial conditions.
(a) ,
Step-by-step calculation:
- (given)
Thus, the sequence alternates between 5 and -5:
(b) ,
Step-by-step calculation:
- (given)
This is an arithmetic progression with a common difference of 3. The general formula is:
(c) ,
Step-by-step calculation:
- (given)
This is a decreasing sequence, and the general solution can be found by summing up the decreasing values of :
(d) ,
Step-by-step calculation:
- (given)
This sequence follows the pattern:
(e) ,
Step-by-step calculation:
- (given)
The sequence grows very quickly. The general formula is:
(f) ,
Step-by-step calculation:
- (given)
The general form for this sequence is:
Would you like more details or have any questions?
Here are some relative questions:
- How do we derive the closed-form solutions for recurrence relations?
- What is the significance of factorials in sequences like part (e)?
- How does the iterative method help in understanding patterns in recurrence relations?
- Can you solve non-linear recurrence relations using similar approaches?
- How would a recurrence relation behave with a different initial condition?
Tip: For recurrence relations, always check if there's a pattern in the first few terms to help you guess a closed-form solution.
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Math Problem Analysis
Mathematical Concepts
Recurrence Relations
Sequences
Iteration
Algebra
Formulas
a_n = (-1)^n * 5
a_n = 1 + 3n
a_n = 4 - (n(n+1))/2
a_n = -2^(n+1) + 1
a_n = 2 * (n+1)!
a_n = 3 * 2^n * n!
Theorems
Arithmetic Progression
Factorial Growth
Suitable Grade Level
Grades 10-12
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