Chapter 12.4, Problem 3AYU

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

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

To determine

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

Answer to Problem 3AYU

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 $3 + 4 + 5 + \cdot \cdot \cdot + (n + 2) = \frac{1}{2}n(n + 5)$ 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 $3 + 4 + 5 + \cdot \cdot \cdot + (n + 2) = \frac{1}{2}n(n + 5)$   -----(1)

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

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

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

That is, to prove that $3 + 4 + 5 + \cdot \cdot \cdot + (k + 3) = \frac{1}{2}(k + 1)(k + 6)$

Consider $3 + 4 + 5 + \cdot \cdot \cdot + (k + 3)$

=$3 + 4 + 5 + \cdot \cdot \cdot + k + 2 + (k + 3)$

$= \frac{1}{2}k(k + 5) + k + 3$ [Substituting equation (1)]

$= \frac{1}{2}[k^2 + 5k + 2k + 6]$

$= \frac{1}{2}[k^2 + 7k + 6] = \frac{1}{2}(k + 1)(k + 6)$

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