Probably use strong induction.Please solve it in a clear way. Better use the format. I have a hard time understanding this question. Please write neatly. Thanks so much Prove that for any positive integer $ n $, there is a $ k \in \mathbb{Z}_{\geq 0} $ and $ a_0, a_1, \ldots, a_k \in \{0, 1\} $ such that\[n = \sum_{i=0}^{k} a_i 2^i.\](For those of you familiar with binary, the equality $ n = \sum_{i=0}^{k} a_i 2^i $ is the same as saying that $ n $ is given by $ a_k a_{k-1} \cdots a_1 a_0 $ in binary. There are several ways to prove this, but one of them uses the fact that given a positive integer $ n $, either $ n = 2m $ for some $ m \in \mathbb{Z} $ or $ n = 2m + 1 $ for some $ m \in \mathbb{Z} $. You may use this fact without proof if you’d like.) **Strong Induction**---### Mathematical Induction Proof1. **Sum Formula:** \[ \sum_{i=1}^{n} i(i+1) = \frac{1}{3}n(n+1)(n+2) \] This formula is verified for all positive integers $ n $ by induction.2. **Proof Explanation:** - Begin with showing the base case. - Assume it holds for $ n = k $. - Prove for $ n = k+1 $.---### Strong Induction Principle#### Axiom 5.4.1 (The Strong Induction Principle)- Assume $ P(n) $ is a statement involving general positive integers $ n $.- **Conditions:** 1. $ P(1) $ is true. 2. For all positive integers $ k $, if $ P(i) $ is true for all positive integers $ i \le k $, then $ P(k+1) $ is true.- **Conclusion:** - Then $ P(n) $ is true for all positive integers $ n $.---**Note:**- Weak induction and strong induction are actually logically equivalent.This explanation is a concise guide to understanding strong induction, emphasizing the principle's logic and its equivalency to weak induction.

Prove that for every positive integer n, there is a k∈ℤ≥0 and a0, a1, …, ak∈0, 1 such that,…

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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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Probably use strong induction. Please solve it in a clear way. Better use the format. I have a hard time understanding this question. Please write neatly. Thanks so much

Prove that for any positive integer $ n $, there is a $ k \in \mathbb{Z}_{\geq 0} $ and $ a_0, a_1, \ldots, a_k \in \{0, 1\} $ such that

\[
n = \sum_{i=0}^{k} a_i 2^i.
\]

(For those of you familiar with binary, the equality $ n = \sum_{i=0}^{k} a_i 2^i $ is the same as saying that $ n $ is given by $ a_k a_{k-1} \cdots a_1 a_0 $ in binary. There are several ways to prove this, but one of them uses the fact that given a positive integer $ n $, either $ n = 2m $ for some $ m \in \mathbb{Z} $ or $ n = 2m + 1 $ for some $ m \in \mathbb{Z} $. You may use this fact without proof if you’d like.)

**Strong Induction**

---

### Mathematical Induction Proof

1. **Sum Formula:**

\[
\sum_{i=1}^{n} i(i+1) = \frac{1}{3}n(n+1)(n+2)
\]

This formula is verified for all positive integers $ n $ by induction.

2. **Proof Explanation:**

- Begin with showing the base case.
- Assume it holds for $ n = k $.
- Prove for $ n = k+1 $.

---

### Strong Induction Principle

#### Axiom 5.4.1 (The Strong Induction Principle)

- Assume $ P(n) $ is a statement involving general positive integers $ n $.

- **Conditions:**
1. $ P(1) $ is true.
2. For all positive integers $ k $, if $ P(i) $ is true for all positive integers $ i \le k $, then $ P(k+1) $ is true.

- **Conclusion:**
- Then $ P(n) $ is true for all positive integers $ n $.

---

**Note:**

- Weak induction and strong induction are actually logically equivalent.

This explanation is a concise guide to understanding strong induction, emphasizing the principle's logic and its equivalency to weak induction.

Expert Solution

Step 1

Prove that for every positive integer $n$ , there is a $k \in ℤ_{\geq 0}$ and $a_{0}, a_{1}, \dots, a_{k} \in \{0, 1\}$ such that,

$n = \sum_{i = 0}^{k} a_{i} 2^{i}$

The objective is to write a natural number as $a_{0} 2^{0} + a_{1} 2^{1} + a_{2} 2^{2} + \dots + a_{k} 2^{k}$ , where, $a_{0}, a_{1}, \dots, a_{k} \in \{0, 1\}$ , that is the coefficients are either $0$ or $1$ . For example,

$\begin{array}{rcl} 3 & = & (1) 2^{0} + (1) 2^{1} \\ 18 & = & (0) 2^{0} + (1) 2^{1} + (0) 2^{2} + (0) 2^{3} + (0) 2^{4} \\ 15 & = & (1) 2^{1} + (1) 2^{1} + (1) 2^{2} + (1) 2^{3} \end{array}$