Question

The terms of a sequence are given recursively as \(p_{0}=3, p_{1}=7,\) and \(p_{n}=3 p_{n-1}-\) \(2 p_{n-2}\) for \(n \geq 2\). Prove by induction that \(b_{n}=2^{n+2}-1\) is a closed form for the sequence.

Short answer

Use strong induction; verify base cases and show the inductive step. Proven: \(b_n = 2^{n+2} - 1\).

Step-by-step solution

Base Case Verification

First, verify the base cases for \(n = 0\) and \(n = 1\). Substitute these into \(b_n = 2^{n+2} - 1\) to check if they match \(p_0\) and \(p_1\). For \(n=0\), \(b_0 = 2^{0+2} - 1 = 4 - 1 = 3\), which matches \(p_0 = 3\). For \(n=1\), \(b_1 = 2^{1+2} - 1 = 8 - 1 = 7\), which matches \(p_1 = 7\). Thus, the base case holds.

Induction Hypothesis

Assume that \(b_k = 2^{k+2} - 1\) holds true for some integer \(k\). Also, assume \(b_{k-1} = 2^{k+1} - 1\). We will show that these assumptions imply \(b_{k+1} = 2^{k+3} - 1\).

Inductive Step

Use the given sequence definition: \(p_n = 3p_{n-1} - 2p_{n-2}\). Substitute\(b_k = 2^{k+2} - 1\) and \(b_{k-1} = 2^{k+1} - 1\) into this equation: \(p_{k+1} = 3b_k - 2b_{k-1}\). Calculate \(3b_k = 3(2^{k+2} - 1) = 3 \times 2^{k+2} - 3\), and \(2b_{k-1} = 2(2^{k+1} - 1) = 2 \times 2^{k+1} - 2\). Subtract: \(3b_k - 2b_{k-1} = 3 \times 2^{k+2} - 3 - (2 \times 2^{k+1} - 2)\).

Simplification

Simplify the expression: \(3 \times 2^{k+2} - 3 - 2 \times 2^{k+1} + 2 = 3 \times 2^{k+2} - 2 \times 2^{k+1} - 1\). Factor out powers of 2: \(= 2^{k+3} - 1\). This matches \(b_{k+1} = 2^{k+3} - 1\).

Conclusion

Since both the base case and the induction step are verified, by the principle of mathematical induction, \(b_n = 2^{n+2} - 1\) is the closed form for the sequence.

Key concepts

Recursive Sequences

Recursive sequences are a way of defining a sequence where each term is derived from previous terms according to a specific rule. Instead of listing out each term explicitly, a recursive rule provides a formula that relates terms to their predecessors. This can make recursive sequences appear complex, but they follow clear patterns.
In the given exercise, the sequence is defined with the initial terms \(p_0 = 3\) and \(p_1 = 7\), and a recursive formula:

• \(p_n = 3p_{n-1} - 2p_{n-2}\) for \(n \geq 2\).

This means that each term from the third onward is determined using the two preceding terms, the product of the previous term scaled by 3, subtracted by twice the second previous term. Recursive sequences help us to comprehend complex systems in a simpler, step-by-step manner.
They are fundamental in mathematical reasoning and often used in computational algorithms.

Closed Form

A closed form is an expression that allows us to compute any term of a sequence directly, without referring back to previous terms, unlike a recursive definition. It is typically in a more explicit and often simplified mathematical form.
In the exercise, the closed form of the sequence is given by

• \(b_n = 2^{n+2} - 1\).

This equation provides a way to find any term of the sequence \(p_n\) solely based on the position \(n\). Once the closed form is verified, it greatly simplifies computing terms in the sequence, especially for larger values of \(n\), since you only need to plug \(n\) into the formula without recalculating every previous sequence term.
Closed forms enhance efficiency in calculations and provide insights into the asymptotic behavior of sequences.

Inductive Hypothesis

The inductive hypothesis is a critical component of a proof by induction. It involves assuming that a statement holds for some arbitrary, fixed natural number \(k\), which allows us to logically show that the same statement must then hold for \(k+1\).
Within the context of the exercise, we assume the statement \(b_k = 2^{k+2} - 1\) holds true for an integer \(k\). By doing this, we can step forward to:

• Prove that if it holds true for \(k\), it should hold for \(k+1\) as well.

This forms the backbone of inductive reasoning, essentially creating a domino effect where if one domino falls (our hypothesis is true for \(k\)), the next one does too (it's true for \(k+1\)), continuing indefinitely. This is a mathematical way of validating that our closed form is correct for all possible terms of \(p_n\).
It's a way of solidifying abstract mathematical relationships into tangible results.

Base Case Verification

Base case verification is the first step in an inductive proof. It's crucial, as it establishes the truth of the statement for the initial value, serving as the "anchor" for the chain of induction.
In the exercise, the base cases are checked for \(n = 0\) and \(n = 1\), verifying:

• \(b_0 = 2^{0+2} - 1 = 3\) matches \(p_0 = 3\), and
• \(b_1 = 2^{1+2} - 1 = 7\) matches \(p_1 = 7\).

By confirming that the closed form holds at these base cases, we ensure that the foundation of our induction is rock solid. Without true base cases, the whole inductive proof would be invalid.
Base case verification is a reminder of the importance of beginnings, both in mathematics and beyond, highlighting the need for accurate initial conditions.