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Assume that \(n\) is a positive integer. Use mathematical induction to prove each statement S by following these steps. See Example \(I\). (a) Verify the statement for \(n=1\) (b) Write the statement for \(n=k\) (c) Write the statement for \(n=k+1\) (d) Assume the statement is true for \(n=k\). Use algebra to change the statement in part (b) to the statement in part (c). (e) Write a conclusion based on Steps (a)-(d). $$2+4+8+\dots+2^{n}=2^{n+1}-2$$

Short Answer

Expert verified
The statement is proved for all positive integers \(n\) by mathematical induction.

Step by step solution

01

Verify the statement for \(n=1\)

The given series is \(2 + 4 + 8 + \dots + 2^n\). For \(n=1\), the left-hand side is \(2\), and the right-hand side is \(2^{1+1} - 2 = 2^2 - 2 = 4 - 2 = 2\). Both sides are equal, so the statement holds for \(n=1\).
02

Write the statement for \(n=k\)

The statement for \(n=k\) is \(2 + 4 + 8 + \dots + 2^k = 2^{k+1} - 2\).
03

Write the statement for \(n=k+1\)

The statement for \(n=k+1\) is \(2 + 4 + 8 + \dots + 2^k + 2^{k+1} = 2^{(k+1)+1} - 2 = 2^{k+2} - 2\).
04

Assume the statement is true for \(n=k\)

Assume the statement is true for \(n=k\), that is, \(2 + 4 + 8 + \dots + 2^k = 2^{k+1} - 2\).
05

Use algebra to change the statement from \(n=k\) to \(n=k+1\)

Starting from the assumption: \[2 + 4 + 8 + \dots + 2^k = 2^{k+1} - 2.\] Add \(2^{k+1}\) to both sides: \[2 + 4 + 8 + \dots + 2^k + 2^{k+1} = 2^{k+1} - 2 + 2^{k+1}.\] Simplify the right-hand side: \[2 + 4 + 8 + \dots + 2^k + 2^{k+1} = 2^{k+1} + 2^{k+1} - 2 = 2 \cdot 2^{k+1} - 2 = 2^{k+2} - 2.\] Thus, the statement holds for \(n=k+1\).
06

Write a conclusion based on Steps 1-5

Since the statement holds for \(n=1\) (base case), and assuming it holds for \(n=k\) leads to it holding for \(n=k+1\) (inductive step), by mathematical induction, the statement \(2 + 4 + 8 + \dots + 2^n = 2^{n+1} - 2\) is true for all positive integers \(n\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Series Summation
Understanding series and their summation is central to many areas of mathematics. In this exercise, we are considering a geometric series where each term is a power of 2. The formula for the summation of the first n terms is given by: \[\sum_{i=1}^n 2^i = 2^{n+1} - 2 \]
What does this mean? It means if you add up the series, each term doubling the previous one, you can find a quick way to get the sum without having to add every single term manually.
For example, summing \{2+4+8+\dots+2^n\} directly would take longer than using the given formula. This formula is derived through recognizing the pattern in the series and makes summation efficient.
Proof by Induction
Proof by induction is a powerful mathematical technique used to prove statements about integers. Let’s break it down into simple steps:
  • Base Case: Verify the statement for the initial value, usually n=1.
  • Inductive Step: Assume the statement is true for some arbitrary positive integer k.
  • Inductive Hypothesis: Use this assumption to prove the statement holds for k+1.

Through these steps, we establish that if the statement is true for one case, and you can prove it leads to the next case being true, the statement must be true for all integers.
In our exercise, we confirmed the series formula for n=1 (base case), then assumed it true for n=k, and showed it must then be true for n=k+1 using algebraic manipulation. This completes the proof by induction.
Algebraic Manipulation
Algebraic manipulation involves using algebraic techniques to rewrite and simplify mathematical expressions or equations. It’s essential for proofs and solving equations.
During the inductive step of our proof, we started with an assumed equation: \{2 + 4 + 8 + \dots + 2^k = 2^{k+1} - 2\}
Then, we added the next term in the series to both sides of the equation:
\[2 + 4 + 8 + \dots + 2^k + 2^{k+1} = 2^{k+1} - 2 + 2^{k+1}\]
We simplified the right-hand side by combining like terms:
\[2^{k+1} + 2^{k+1} - 2 = 2\cdot2^{k+1} - 2 = 2^{k+2} - 2\]
This shows that the statement holds for n=k+1 if it holds for n=k. Mastery of algebraic manipulation helps make such logical transitions clear and confirms the validity of mathematical statements.

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