Chapter 12.4, Problem 11AYU

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

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

To determine

To prove: The given statement $\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + \cdot \cdot \cdot + \frac{1}{n(n + 1)} = \frac{n}{n + 1}$ is true for all natural numbers $n$ using the Principle of Mathematical Induction.

Answer to Problem 11AYU

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 $\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + \cdot \cdot \cdot + \frac{1}{n(n + 1)} = \frac{n}{n + 1}$ 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 $\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + \cdot \cdot \cdot + \frac{1}{n(n + 1)} = \frac{n}{n + 1}$   -----(1)

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

That is $\frac{1}{1.2} = \frac{1}{1 + 1}.$ 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 $\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + \cdot \cdot \cdot + \frac{1}{k(k + 1)} = \frac{k}{k + 1}$   -----(1)

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

That is, to prove that $\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + \cdot \cdot \cdot + \frac{1}{(k + 1)(k + 1 + 1)} = \frac{k + 1}{k + 1 + 1}$

$\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + \cdot \cdot \cdot + \frac{1}{k(k + 1)} + \frac{1}{(k + 1)(k + 2)} = \frac{k + 1}{k + 2}$

Consider LHS $\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + \cdot \cdot \cdot + \frac{1}{k(k + 1)} + \frac{1}{(k + 1)(k + 2)}$

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

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

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