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) an=an1a_n = -a_{n-1}, a0=5a_0 = 5

Step-by-step calculation:

  • a0=5a_0 = 5 (given)
  • a1=a0=5a_1 = -a_0 = -5
  • a2=a1=5a_2 = -a_1 = 5
  • a3=a2=5a_3 = -a_2 = -5
  • a4=a3=5a_4 = -a_3 = 5

Thus, the sequence alternates between 5 and -5: an=(1)n5a_n = (-1)^n \cdot 5


(b) an=an1+3a_n = a_{n-1} + 3, a0=1a_0 = 1

Step-by-step calculation:

  • a0=1a_0 = 1 (given)
  • a1=a0+3=1+3=4a_1 = a_0 + 3 = 1 + 3 = 4
  • a2=a1+3=4+3=7a_2 = a_1 + 3 = 4 + 3 = 7
  • a3=a2+3=7+3=10a_3 = a_2 + 3 = 7 + 3 = 10
  • a4=a3+3=10+3=13a_4 = a_3 + 3 = 10 + 3 = 13

This is an arithmetic progression with a common difference of 3. The general formula is: an=1+3na_n = 1 + 3n


(c) an=an1na_n = a_{n-1} - n, a0=4a_0 = 4

Step-by-step calculation:

  • a0=4a_0 = 4 (given)
  • a1=a01=41=3a_1 = a_0 - 1 = 4 - 1 = 3
  • a2=a12=32=1a_2 = a_1 - 2 = 3 - 2 = 1
  • a3=a23=13=2a_3 = a_2 - 3 = 1 - 3 = -2
  • a4=a34=24=6a_4 = a_3 - 4 = -2 - 4 = -6

This is a decreasing sequence, and the general solution can be found by summing up the decreasing values of nn: an=4n(n+1)2a_n = 4 - \frac{n(n+1)}{2}


(d) an=2an13a_n = 2a_{n-1} - 3, a0=1a_0 = -1

Step-by-step calculation:

  • a0=1a_0 = -1 (given)
  • a1=2a03=2(1)3=23=5a_1 = 2a_0 - 3 = 2(-1) - 3 = -2 - 3 = -5
  • a2=2a13=2(5)3=103=13a_2 = 2a_1 - 3 = 2(-5) - 3 = -10 - 3 = -13
  • a3=2a23=2(13)3=263=29a_3 = 2a_2 - 3 = 2(-13) - 3 = -26 - 3 = -29
  • a4=2a33=2(29)3=583=61a_4 = 2a_3 - 3 = 2(-29) - 3 = -58 - 3 = -61

This sequence follows the pattern: an=2n+1+1a_n = -2^{n+1} + 1


(e) an=(n+1)an1a_n = (n+1)a_{n-1}, a0=2a_0 = 2

Step-by-step calculation:

  • a0=2a_0 = 2 (given)
  • a1=(1+1)a0=22=4a_1 = (1+1) \cdot a_0 = 2 \cdot 2 = 4
  • a2=(2+1)a1=34=12a_2 = (2+1) \cdot a_1 = 3 \cdot 4 = 12
  • a3=(3+1)a2=412=48a_3 = (3+1) \cdot a_2 = 4 \cdot 12 = 48
  • a4=(4+1)a3=548=240a_4 = (4+1) \cdot a_3 = 5 \cdot 48 = 240

The sequence grows very quickly. The general formula is: an=2(n+1)!a_n = 2 \cdot (n+1)!


(f) an=2nan1a_n = 2na_{n-1}, a0=3a_0 = 3

Step-by-step calculation:

  • a0=3a_0 = 3 (given)
  • a1=21a0=213=6a_1 = 2 \cdot 1 \cdot a_0 = 2 \cdot 1 \cdot 3 = 6
  • a2=22a1=226=24a_2 = 2 \cdot 2 \cdot a_1 = 2 \cdot 2 \cdot 6 = 24
  • a3=23a2=2324=144a_3 = 2 \cdot 3 \cdot a_2 = 2 \cdot 3 \cdot 24 = 144
  • a4=24a3=24144=1152a_4 = 2 \cdot 4 \cdot a_3 = 2 \cdot 4 \cdot 144 = 1152

The general form for this sequence is: an=32nn!a_n = 3 \cdot 2^n \cdot n!


Would you like more details or have any questions?
Here are some relative questions:

  1. How do we derive the closed-form solutions for recurrence relations?
  2. What is the significance of factorials in sequences like part (e)?
  3. How does the iterative method help in understanding patterns in recurrence relations?
  4. Can you solve non-linear recurrence relations using similar approaches?
  5. 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