Chapter 12.4, Problem 16AYU

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

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

To determine

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

Answer to Problem 16AYU

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

Explanation of Solution

Given:

Statements says the series $-2 - 3 - 4- \cdots -\left(n + 1\right) = -\frac{1}{2} n\left(n + 3\right)$ 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 $-2 - 3 - 4- \cdots -\left(n + 1\right) = -\frac{1}{2} n\left(n + 3\right)$   -----(1)

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

That is $-\left(1 + 1\right) = -\frac{1}{2}.1. \left(1 + 3\right).$ 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 $-2 - 3 - 4- \cdots -\left(k + 1\right) = -\frac{1}{2} k\left(k + 3\right)$   -----(1)

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

That is, to prove that

$-2 - 3 - 4- \cdots -\left(k + 1\right)-\left(k + 1 + 1\right) = -\frac{1}{2}\left(k + 1\right)\left(k + 1 + 3\right)$

$-2-3-4-\cdots -\left(k+1\right)-\left(k+2\right)=-\frac{1}{2}\left(k+1\right)\left(k+4\right)$

Consider $\text{LHS}-2 - 3 - 4- \cdots -\left(k + 1\right)-\left(k + 2\right) = -\frac{1}{2} k\left(k + 3\right)-k + 2$

$= -\frac{1}{2} \left(k^2 + 3k + 2k + 4\right) = -\frac{1}{2} \left(k^2 + 5k + 4\right) = -\frac{1}{2}\left(k + 1\right)\left(k + 4\right)$

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

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