Chapter 12.4, Problem 9AYU

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

$1{\text{ + 4 + 4}}^{\text{2}}\text{ + }\mathrm{...}{\text{ + 4}}^{n-\text{ 1}}\text{ = }\frac{1}{3}\left(4^n-\text{ 1}\right)$

To determine

To prove: The given statement $1 + 4 + 4^2 + \cdot \cdot \cdot + 4^{n - 1} = \frac{1}{3}(4^n - 1)$ is true for all natural numbers $n$ using the Principle of Mathematical Induction.

Answer to Problem 9AYU

As the statement is true for the natural number $(k + 1)$ terms, hence the statement is true for all natural numbers.

Explanation of Solution

Given:

Statements says the series $1 + 4 + 4^2 + \cdot \cdot \cdot + 4^{n - 1} = \frac{1}{3}(4^n - 1)$ is true for all natural number.

Formula used:

The Principle of Mathematical Induction

Suppose that the following two conditions are satisfied with regard to a statement about natural numbers:

CONDITION I: The statement is true for the natural number 1.

CONDITION II: If the statement is true for some natural number $k$, it is also true for the next natural number $k + 1$. Then the statement is true for all natural numbers.

Proof:

Consider the statement $1 + 4 + 4^2 + \cdot \cdot \cdot + 4^{n - 1} = \frac{1}{3}(4^n - 1)$   -----(1)

Step 1: Show that statement (1) is true for $n = 1$.

That is $4^{1 - 1} = \frac{1}{3}(4^1 - 1).$ Hence the statement is true for natural number $n = 1$.

Step 2: Assume that the statement is true for some natural number $k$.

That is $1 + 4 + 4^2 + \cdot \cdot \cdot + 4^{k - 1} = \frac{1}{3}(4^k - 1)$   -----(1)

Step 3: Prove that the statement is true for the next natural number $k + 1$.

That is, to prove that

$1 + 4 + 4^2 + \cdot \cdot \cdot + 4^{(k + 1) - 1} = \frac{1}{3}(4^{k + 1} - 1)$

Consider LHS $1 + 4 + 4^2 + \cdot \cdot \cdot + 4^{k - 1} + 4^{(k + 1) - 1}$

$= 1 + 4 + 4^2 + \cdot \cdot \cdot + 4^{k - 1} + 4^k = \frac{1}{3}{(4}^k - 1) + 4^k = \frac{4}{3}.4^k - \frac{1}{3} = \frac{1}{3}{(4}^{k + 1} - 1)$

$\text{LHS}=\text{RHS}$

As the statement is true for the natural number $(k + 1)$ terms, hence the statement is true for all natural numbers.