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4.34 con't. and so a+²= ax+1-1 +29+1-2 акн - аки- = a + 2a₂²-1 = 2K-¹ + 2¹ (2-2) = 2k-1 +2k-1 H 2(2k-1) 2(k+1)-1 which is the desired result. By the Strong Principle of Mathematical Induction 20-1 Ap all DEN. for 2

4.34 con't. and so a+²= ax+1-1 +29+1-2 акн - аки- = a + 2a₂²-1 = 2K-¹ + 2¹ (2-2) = 2k-1 +2k-1 H 2(2k-1) 2(k+1)-1 which is the desired result. By the Strong Principle of Mathematical Induction 20-1 Ap all DEN. for 2

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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May I get a proof analysis? 

**Title: Recursive Sequence and Inductive Proof** ### Problem Statement A sequence \(\{a_n\}\) is defined recursively by: - \(a_1 = 1\), - \(a_2 = 2\), - \(a_n = a_{n-1} + 2a_{n-2}\) for \(n \geq 3\). Conjecture a formula for \(a_n\) and verify it. ### Initial Calculations - \(a_2 = a_1 + 2a_0 = 2\), - \(a_3 = a_2 + 2a_1 = 4\), - \(a_4 = a_3 + 2a_2 = 8 = 2^{4-1}\), - \(a_5 = a_4 + 2a_3 = 16\), - \(a_6 = a_5 + 2a_4 = 32 = 2^{6-1}\). ### Conjecture The sequence \(\{a_n\}\) is defined recursively by: - \(a_1 = 1\), - \(a_2 = 2\), - \(a_n = a_{n-1} + 2a_{n-2}\) for \(n \geq 3\). The conjecture is: \(a_n = 2^{n-1}\) for all \(n \in \mathbb{N}\). ### Proof by Induction **Base Case**: Verify for \(n = 1\): - \(a_1 = 2^{1-1} = 2^0 = 1\). The formula holds for \(n = 1\). **Inductive Step**: Assume for an arbitrary positive integer \(k\) that \(a_k = 2^{k-1}\). We want to show that \(a_{k+1} = 2^{(k+1)-1} = 2^k\). - Since \(a_2 = 8\), it follows that \(a_{k+1} = 2^{(k+1)-1}\) when \(k = 3\). Thus, we may assume that \(k = 3\). Hence, \(k+1 \leq 4\). This completes the proof by
Transcribed Image Text:**Title: Recursive Sequence and Inductive Proof** ### Problem Statement A sequence \(\{a_n\}\) is defined recursively by: - \(a_1 = 1\), - \(a_2 = 2\), - \(a_n = a_{n-1} + 2a_{n-2}\) for \(n \geq 3\). Conjecture a formula for \(a_n\) and verify it. ### Initial Calculations - \(a_2 = a_1 + 2a_0 = 2\), - \(a_3 = a_2 + 2a_1 = 4\), - \(a_4 = a_3 + 2a_2 = 8 = 2^{4-1}\), - \(a_5 = a_4 + 2a_3 = 16\), - \(a_6 = a_5 + 2a_4 = 32 = 2^{6-1}\). ### Conjecture The sequence \(\{a_n\}\) is defined recursively by: - \(a_1 = 1\), - \(a_2 = 2\), - \(a_n = a_{n-1} + 2a_{n-2}\) for \(n \geq 3\). The conjecture is: \(a_n = 2^{n-1}\) for all \(n \in \mathbb{N}\). ### Proof by Induction **Base Case**: Verify for \(n = 1\): - \(a_1 = 2^{1-1} = 2^0 = 1\). The formula holds for \(n = 1\). **Inductive Step**: Assume for an arbitrary positive integer \(k\) that \(a_k = 2^{k-1}\). We want to show that \(a_{k+1} = 2^{(k+1)-1} = 2^k\). - Since \(a_2 = 8\), it follows that \(a_{k+1} = 2^{(k+1)-1}\) when \(k = 3\). Thus, we may assume that \(k = 3\). Hence, \(k+1 \leq 4\). This completes the proof by
Certainly! Here’s the transcription for the educational website: --- **6.34 cont.** And so, \[ a_{k+1} = a_{k+1} + 2a_{k+1-2} \] \[ = a_{k} + 2a_{k-1} \] \[ = 2^{k-1} + 2(2^{k-2}) \] \[ = 2^{k-1} + 2^{k-1} \] \[ = 2(2^{k-1}) \] \[ = 2^{(k+1)-1} \] which is the desired result. By the Strong Principle of Mathematical Induction, \( a_{n} = 2^{n-1} \) for all \( n \in \mathbb{N} \). --- **Explanation:** This section demonstrates a mathematical proof using the Strong Principle of Mathematical Induction. It shows a step-by-step derivation to establish that the expression \( a_n = 2^{n-1} \) holds true for all natural numbers \( n \). The calculations illustrate how the sequence \( a_n \) can be expressed as a function of powers of 2, reinforcing its correctness through algebraic manipulations.
Transcribed Image Text:Certainly! Here’s the transcription for the educational website: --- **6.34 cont.** And so, \[ a_{k+1} = a_{k+1} + 2a_{k+1-2} \] \[ = a_{k} + 2a_{k-1} \] \[ = 2^{k-1} + 2(2^{k-2}) \] \[ = 2^{k-1} + 2^{k-1} \] \[ = 2(2^{k-1}) \] \[ = 2^{(k+1)-1} \] which is the desired result. By the Strong Principle of Mathematical Induction, \( a_{n} = 2^{n-1} \) for all \( n \in \mathbb{N} \). --- **Explanation:** This section demonstrates a mathematical proof using the Strong Principle of Mathematical Induction. It shows a step-by-step derivation to establish that the expression \( a_n = 2^{n-1} \) holds true for all natural numbers \( n \). The calculations illustrate how the sequence \( a_n \) can be expressed as a function of powers of 2, reinforcing its correctness through algebraic manipulations.
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