For every integer n23 I+ 2+ 2?+ +4^= 4(4n-16) %3D 3. Proof (by induction)

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ISBN:9780470458365
Author:Erwin Kreyszig
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**Mathematical Induction Proof for a Summation Formula**

This document presents a proof by induction for a mathematical statement that holds for every integer \( n \geq 3 \).

**Statement:**

For every integer \( n \geq 3 \), the sum:

\[ 1 + 2 + 2^2 + \ldots + 4^n \]

is equal to:

\[ \frac{4(4^n - 16)}{3} \]

**Proof Outline**:

The proof is likely structured as a mathematical induction. Here's a typical approach:

1. **Base Case**: Verify the statement for the initial integer \( n = 3 \).
2. **Inductive Step**: Assume the statement holds for an arbitrary integer \( n = k \). Then prove it also holds for \( n = k + 1 \).
3. **Conclusion**: Conclude that the statement is true for all integers \( n \geq 3 \) based on the base case and inductive step.

This foundational method shows how each subsequent step can be deduced from the previous one, establishing the general validity of the formula.
Transcribed Image Text:**Mathematical Induction Proof for a Summation Formula** This document presents a proof by induction for a mathematical statement that holds for every integer \( n \geq 3 \). **Statement:** For every integer \( n \geq 3 \), the sum: \[ 1 + 2 + 2^2 + \ldots + 4^n \] is equal to: \[ \frac{4(4^n - 16)}{3} \] **Proof Outline**: The proof is likely structured as a mathematical induction. Here's a typical approach: 1. **Base Case**: Verify the statement for the initial integer \( n = 3 \). 2. **Inductive Step**: Assume the statement holds for an arbitrary integer \( n = k \). Then prove it also holds for \( n = k + 1 \). 3. **Conclusion**: Conclude that the statement is true for all integers \( n \geq 3 \) based on the base case and inductive step. This foundational method shows how each subsequent step can be deduced from the previous one, establishing the general validity of the formula.
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