Chapter 12.4, Problem 2AYU

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

$1\text{ + 5 + 9 + }\mathrm{...}\text{ + }\left(4n-\text{ 3}\right)\text{ = }n\left(2n-\text{ 1}\right)$

To determine

To prove: The given statement $(1 + 5 + 9 + \cdot \cdot \cdot + )4n - 3 = n(2n - 1)$ is true for all natural numbers $n$ using the Principle of Mathematical Induction.

Answer to Problem 2AYU

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 $(1 + 5 + 9 + \cdot \cdot \cdot + )4n - 3 = n(2n - 1)$ 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 $(1 + 5 + 9 + \cdot \cdot \cdot + )4n - 3 = n(2n - 1)$   -----(1)

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

That is $4(1) - 3 = 1(2(1) - 1).$ 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 $(1 + 5 + 9 + \cdot \cdot \cdot + )4k - 3 = k(2k - 1)$   -----(1)

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

That is, to prove that $(1 + 5 + 9 + \cdot \cdot \cdot + 4(k + 1) - 3) = (k + 1)(2(k + 1) - 1) = (k + 1)(2k + 1)$

Consider $2 + 4 + 6 + \cdot \cdot \cdot + 4k + 1$

=$2 + 4 + 6 + \cdot \cdot \cdot + 4k - 1 + 4k + 1$

$= k(2k - 1) + 4k + 1$ [Substituting equation (1)]

$= 2k^2 - k + 4k + 1 = 2k^2 + 3k + 1 = (k + 1)(2k + 1)$

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