Math Problem Statement

Suppose that a sequence is defined by

a1 = 3, an + 1  = 
1 6 − an  .

To show that an is monotonic using mathematical induction, which of the following would be the second step? (A) Assume  ak + 1  ≤  ak  and show that   1 6 − ak + 1  ≤ 
1 6 − ak

(B) Assume   1 6 − ak  ≤ 
1 6 − ak + 1   and show that   ak  ≤  ak + 1 (C) Assume   1 6 − ak + 1  ≤ 
1 6 − ak   and show that   ak + 1  ≤  ak (D) Assume  ak + 1  ≤  ak  and show that   1 6 − ak + 2  ≤ 
1 6 − ak + 1

(E) Assume   1 6 − ak + 2  ≤ 
1 6 − ak + 1   and show that   ak + 1  ≤  ak (F) Assume  ak  ≤  ak + 1  and show that   1 6 − ak + 1  ≤ 
1 6 − ak + 2

(G) Assume  ak  ≤  ak + 1  and show that   1 6 − ak  ≤ 
1 6 − ak + 1

(H) Assume   1 6 − ak + 1  ≤ 
1 6 − ak + 2   and show that   ak  ≤  ak + 1

Solution

To prove that the sequence ana_n is monotonic using mathematical induction, we need to show that the sequence is either increasing or decreasing. We can do this by proving that if aka_k satisfies the monotonicity condition, then ak+1a_{k+1} continues to satisfy the same condition.

The recursive relation for the sequence is:

a1=3,an+1=16ana_1 = 3, \quad a_{n+1} = \frac{1}{6 - a_n}

Step-by-step reasoning:

  1. Base Case: You will first show that the sequence is either increasing or decreasing for a1a_1.

  2. Inductive Step: For the inductive step, you assume that aka_k satisfies the condition of being either greater than or less than ak+1a_{k+1} (depending on whether the sequence is increasing or decreasing).

  3. Then, you need to show that ak+1a_{k+1} continues to satisfy this condition for ak+2a_{k+2}.

Analyzing the choices:

  • Option (A): Assuming ak+1aka_k + 1 \leq a_k and showing that 16ak+116ak\frac{1}{6 - a_{k+1}} \leq \frac{1}{6 - a_k} is useful if you are assuming the sequence is decreasing.

  • Option (B): Assuming 16ak16ak+1\frac{1}{6 - a_k} \leq \frac{1}{6 - a_{k+1}} and showing that akak+1a_k \leq a_{k+1} is useful if you are assuming the sequence is increasing.

  • Option (C): This would assume the sequence is decreasing but reverses the inequality for aka_k and ak+1a_{k+1}. It seems like the wrong direction for a correct inductive step.

Thus, Option (B) is the best choice, because if you assume 16ak16ak+1\frac{1}{6 - a_k} \leq \frac{1}{6 - a_{k+1}}, it directly leads to proving the sequence is increasing.


Final Answer:

(B) Assume 16ak16ak+1\frac{1}{6 - a_k} \leq \frac{1}{6 - a_{k+1}} and show that akak+1a_k \leq a_{k+1}.


Would you like a detailed explanation of the full induction proof or more clarification on the steps?

Follow-up questions:

  1. How do you prove the base case in induction for sequences?
  2. What does it mean for a sequence to be monotonic?
  3. How can you generalize induction for different types of sequence relations?
  4. What are some common errors when applying mathematical induction to sequences?
  5. How can the recursive formula an+1=16ana_{n+1} = \frac{1}{6 - a_n} be analyzed for convergence?

Tip: When working with recursive sequences, try analyzing the behavior of the recursive formula by considering special cases (like a1a_1, a2a_2, etc.) to identify patterns or trends early in the proof.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Sequences
Monotonicity
Mathematical Induction
Recursive Relations

Formulas

a1 = 3
a_{n+1} = 1 / (6 - a_n)

Theorems

Mathematical Induction

Suitable Grade Level

Grades 11-12