Chapter 12.4, Problem 10AYU

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

$1{\text{ + 5 + 5}}^{\text{2}}\text{ + }\mathrm{...}{\text{ + 5}}^{n-\text{ 1}}\text{ = }\frac{1}{4}\left(5^n-\text{ 1}\right)$

To determine

To prove: The given statement $1 + 5 + 5^2 + \cdot \cdot \cdot + 5^{n - 1} = \frac{1}{4}(5^n - 1)$ is true for all natural numbers $n$ using the Principle of Mathematical Induction.

Answer to Problem 10AYU

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 $1 + 5 + 5^2 + \cdot \cdot \cdot + 5^{n - 1} = \frac{1}{4}(5^n - 1)$ 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 $1 + 5 + 5^2 + \cdot \cdot \cdot + 5^{n - 1} = \frac{1}{4}(5^n - 1)$   -----(1)

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

That is $5^{1 - 1} = \frac{1}{4}(5^1 - 1).$ 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 $1 + 5 + 5^2 + \cdot \cdot \cdot + 5^{k - 1} = \frac{1}{4}(5^k - 1)$   -----(1)

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

That is, to prove that

$1 + 5 + 5^2 + \cdot \cdot \cdot + 5^{(k + 1) - 1} = \frac{1}{4}(5^{k + 1} - 1)$

Consider LHS $1 + 5 + 5^2 + \cdot \cdot \cdot + 5^{k - 1} + 5^{(k + 1) - 1}$

$= 1 + 5 + 5^2 + \cdot \cdot \cdot + 5^{k - 1} + 5^k = \frac{1}{4}{(5}^k - 1) + 5^k = \frac{5}{4}.5^k - \frac{1}{4} = \frac{1}{4}{(5}^{k + 1} - 1)$

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

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