Chapter 12.4, Problem 6AYU

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

$1\text{ + 4 + 7 + }\mathrm{...}\text{ + }\left(3n-\text{ 2}\right)\text{ = }\frac{1}{2}n\left(3n-\text{ 1}\right)$

To determine

To prove: The given statement $1 + 4 + 7 + \cdot \cdot \cdot + (3n - 2) = \frac{1}{2}n(3n - 1)$ is true for all natural numbers $n$ using the Principle of Mathematical Induction.

Answer to Problem 6AYU

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 $1 + 4 + 7 + \cdot \cdot \cdot + (3n - 2) = \frac{1}{2}n(3n - 1)$ 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 $\mathrm{k}$, it is also true for the next natural number $k + 1$. Then the statement is true for all natural numbers.

Proof:

Consider the statement $1 + 4 + 7 + \cdot \cdot \cdot + (3n - 2) = \frac{1}{2}n(3n - 1)$   -----(1)

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

That is $3(1) - 2 = \frac{1}{2}.1.(3(1) - 1).$ 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 $1 + 4 + 7 + \cdot \cdot \cdot + (3k - 2) = \frac{1}{2}k(3k - 1)$   -----(1)

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

That is, to prove that

$1 + 4 + 7 + \cdot \cdot \cdot + (3k - 2) + (3(k + 1) - 2) = \frac{1}{2}(k + 1)(3(k + 1) - 1)$

$1 + 4 + 7 + \cdot \cdot \cdot + (3k - 2) + (3k + 1) = \frac{1}{2}(k + 1)(3k + 2)$

Consider LHS $1 + 4 + 7 + \cdot \cdot \cdot + (3k + 1) + (3k + 1) = \frac{1}{2}k(3k - 1) + (3k + 1)$

$= \frac{1}{2}(3k^2 + 5k + 2)$

Consider RHS $\frac{1}{2}(k + 1)(3k + 2) = \frac{1}{2}(3k^2 + 2k + 3k + 2) = \frac{1}{2}(3k^2 + 5k + 2).$

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

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