Question

Given that \(b_{n-1}=2^{n+1}-1\) and \(b_{n-2}=2^{n}-1,\) prove that if \(b_{n}=3 b_{n-1}-2 b_{n-2}\) then \(b_{n}=2^{n+2}-1\) provided \(n \geq 2\).

Short answer

The expression for \(b_n\) satisfies \(b_n = 2^{n+2} - 1\), confirming the hypothesis for \(n \geq 2\).

Step-by-step solution

Understand the Problem

The problem gives us definitions for \(b_{n-1}\) and \(b_{n-2}\) and asks us to prove that their relationship with \(b_n\) defined as \(b_n = 3b_{n-1} - 2b_{n-2}\) leads to \(b_n = 2^{n+2} - 1\) for \(n \geq 2\).

Substitute Expressions for Previous Terms

Substitute the given expressions for \(b_{n-1}\) and \(b_{n-2}\) into the equation for \(b_n\). We replace \(b_{n-1}\) with \(2^{n+1} - 1\) and \(b_{n-2}\) with \(2^n - 1\): \[ b_n = 3(2^{n+1} - 1) - 2(2^n - 1) \]

Distribute the Constants in the Equation

Distribute the constants in the equation from Step 2:\[ b_n = 3 \times 2^{n+1} - 3 - 2 \times 2^n + 2 \]

Simplify the Expression

Combine like terms:\[ b_n = 3 \times 2^{n+1} - 2 \times 2^n + (2 - 3) \]Simplify the constants:\[ b_n = 3 \times 2^{n+1} - 2 \times 2^n - 1 \]

Factor and Further Simplify the Expression

Recognize that \(3 \times 2^{n+1} = 3 \times (2 \times 2^n) = 6 \times 2^n\). Thus, rewrite and simplify further:\[ b_n = 6 \times 2^n - 2 \times 2^n - 1 \] Now factor out \(2^n\):\[ b_n = (6 - 2) \times 2^n - 1 = 4 \times 2^n - 1 \]

Final Expression Matches the Form

Recognize that \(4 \times 2^n = 2^{n+2}\). Therefore, replace it in the expression:\[ b_n = 2^{n+2} - 1 \]This matches the hypothesis \(b_n = 2^{n+2} - 1\), confirming that the given condition holds.

Key concepts

Mathematical Induction

Mathematical induction is a powerful technique used to prove statements that are formulated for all natural numbers. It operates much like dominoes falling; you show one falls (base case) and each one causes the next to fall (inductive step). In our problem, once we prove the base case and show that if the statement holds for one number it holds for the next, then the statement is true for all numbers greater than or equal to 2.

Here's a step-by-step view of how mathematical induction works:

• Base Case: Prove the statement is true for the first value that the variable can take, typically when \(n=1\) or another appropriate starting point based on the problem.
• Inductive Step: Assume the statement is true for some arbitrary natural number \(k\). Then prove the statement holds for \(k+1\).

This structured method ensures the conclusion applies to all subsequent values.It's a rigorous approach, which offers certainty in areas like discrete mathematics, where preciseness is essential.

Discrete Mathematics

Discrete mathematics is the branch of mathematics dealing with counts that are distinct and separate. It contrasts with continuous mathematics, like calculus, which deals with limits and infinitesimally small sizes. Discrete mathematics focuses on integers, graphs, logical statements, and other distinct elements.

Key concepts often included are:

• Sequences and series, like the one in our problem.
• Graph theory and networks.
• Logic and set theory.
• Combinatorics—the art of counting strategically.

In our exercise, the use of discrete sequences and recurrence relations highlights how discrete mathematics allows us to break problems into a set of distinct, related parts. By understanding each part's role, we navigate towards a comprehensive solution.

Sequence Proof

Proving statements about sequences involves showing that a given expression for the sequence holds for all terms in the sequence. In our problem, we want to prove the recursive sequence formula holds true.
To establish this, we follow a logical path:

• Identify a pattern or rule in a sequence that needs proving, such as our recurrence relation \(b_n = 3b_{n-1} - 2b_{n-2}\).
• Substitute known values or terms into the sequence formula, thereby reducing complexity and focusing on confirmation.
• Simplify the expression to reveal consistency with the desired formula \(b_n = 2^{n+2} - 1\). This ensures that the pattern matches the generic rule for all sequence numbers.

This approach allows us to systematically validate whether an expression accurately represents a sequence, maintaining consistency throughout all elements.