Chapter 12: Q. 7 (page 830)

In Problems 1–22, use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers $n .$
$1 + 2 + 2^{2} + . . . + 2^{n - 1} = 2^{n} - 1$

Short Answer

Expert verified

The statement is true for all natural numbers.

Step by step solution

Step 1. Given information

The given statement is $1 + 2 + 2^{2} + . . . + 2^{n - 1} = 2^{n} - 1 .$ We need to prove that the given statement is true for all natural numbers.

Step 2. Proved

First, we show that the statement is true when $n = 1 .$

$1 = 2^{1} - 1 .$

$= 1 .$

Condition $1$ of the principle of mathematical induction holds.

Let's assume that the statement is true for $n = k .$

So, $1 + 2 + 2^{2} + . . . + 2^{k - 1} = 2^{k} - 1 \dots (1) .$

Let's prove that the statement is true for $n = k + 1 .$

localid="1648053650425" $1 + 2 + 2^{2} + . . . + 2^{k - 1} + 2^{k + 1 - 1} .$

$= 2^{k} - 1 + 2^{k} .$

$= 2 \cdot 2^{k} - 1 .$

$= 2^{k + 1} - 1 .$

Since the given statement is true for $n = 1$ and if the statement is true for some $k$ and it is also true for the next natural number $k + 1 .$

So, the statement is true for all natural numbers.

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In Problems 1–22, use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers $n .$
$1 + 2 + 2^{2} + . . . + 2^{n - 1} = 2^{n} - 1$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Proved

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In Problems 1–22, use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers n.1+2+22+...+2n-1=2n-1

Short Answer

Step by step solution

Step 1. Given information

Step 2. Proved

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Applied Mathematics

Calculus

Pure Maths

Logic and Functions

Probability and Statistics

Study anywhere. Anytime. Across all devices.

In Problems 1–22, use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers $n .$
$1 + 2 + 2^{2} + . . . + 2^{n - 1} = 2^{n} - 1$