Chapter 12.4, Problem 12AYU

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

$\frac{1}{1\cdot 3}\text{ + }\frac{1}{3\cdot 5}\text{ + }\frac{1}{5\cdot 7}\text{ + }\mathrm{...}\text{ + }\frac{1}{\left(2n-\text{ 1}\right)\left(2n+\text{ 1}\right)}\text{ = }\frac{n}{2n+1}$

To determine

To prove: The given statement $\frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + \cdot \cdot \cdot + \frac{1}{(2n - 1)(2n + 1)} = \frac{n}{2n + 1}$ is true for all natural numbers $n$ using the Principle of Mathematical Induction.

Answer to Problem 12AYU

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 $\frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + \cdot \cdot \cdot + \frac{1}{(2n - 1)(2n + 1)} = \frac{n}{2n + 1}$ 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 $\frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + \cdot \cdot \cdot + \frac{1}{(2n - 1)(2n + 1)} = \frac{n}{2n + 1}$   -----(1)

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

That is $\frac{1}{(2 - 1).(2 + 1)} = \frac{1}{2 + 1}.$ 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 $\frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + \cdot \cdot \cdot + \frac{1}{(2k - 1)(2k + 1)} = \frac{k}{2k + 1}$   -----(1)

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

That is, to prove that

$\frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + \cdot \cdot \cdot + \frac{1}{(2k - 1)(2k + 1)} + \frac{1}{(2(k + 1) - 1)(2(k + 1) + 1)} = \frac{k + 1}{2(k + 1) + 1}$

$\frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + \cdot \cdot \cdot + \frac{1}{(2k - 1)(2k + 1)} + \frac{1}{(2k + 1)(2k + 3)} = \frac{k + 1}{2k + 3}$

Consider LHS $\frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + \cdot \cdot \cdot + \frac{1}{(2k - 1)(2k + 1)} + \frac{1}{(2k + 1)(2k + 3)}$

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

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

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