Question

Use mathematical induction to show that

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

whenever nis a positive integer.

Short answer

It is shown that\(\frac{1}{{1 \cdot 3}} + \frac{1}{{3 \cdot 5}} + ... + \frac{1}{{\left( {2n - 1} \right) \cdot \left( {2n + 1} \right)}} = \frac{n}{{2n + 1}}\)whenever\(n\)is a positive integer.

Step-by-step solution

Principle of Mathematical Induction

Consider the propositional function\(P\left( n \right)\). Consider two actions to prove that\(P\left( n \right)\) evaluates to accurate for all set of positive integers\(n\).

Consider the first basic step is to confirm that\(P\left( 1 \right)\)true.

Consider the inductive step is to demonstrate that for any positive integer k the conditional statement\(P\left( k \right) \to P\left( {k + 1} \right)\)is true.

Prove the basis step

Given statement is:

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

In the basis step, we need to prove that\(P\left( 1 \right)\)is true.

For finding statement\(P\left( 1 \right)\)substituting\(1\)for\(n\)in the statement.

Therefore, the statement\(P\left( 1 \right)\)is:

\(\begin{array}{c}\frac{1}{{1 \cdot 3}} = \frac{1}{{2\left( 1 \right) + 1}}\\\frac{1}{3} = \frac{1}{3}\end{array}\)

Therefore, the statement \(P\left( 1 \right)\) is true this is also known as the basis step of the proof.

Prove the Inductive step

In the inductive step, we need to prove that, if\(P\left( k \right)\)is true, then\(P\left( {k + 1} \right)\)is also true.

That is,

\(P\left( k \right) \to P\left( {k + 1} \right)\)is true for all positive integers k.

In the inductive hypothesis, we assume that\(P\left( k \right)\)is true for any arbitrary positive integer\(k\)

That is

\(\frac{1}{{1 \cdot 3}} + \frac{1}{{3 \cdot 5}} + ... + \frac{1}{{\left( {2k - 1} \right) \cdot \left( {2k + 1} \right)}} = \frac{k}{{2k + 1}}\)...(i)

Now we must have to show that\(P\left( {k + 1} \right)\)is also true.

Therefore replacing\(k\)with\(k + 1\)in the statement

\(\frac{1}{{1 \cdot 3}} + \frac{1}{{3 \cdot 5}} + ... + \frac{1}{{\left( {2\left( {k + 1} \right) - 1} \right) \cdot \left( {2\left( {k + 1} \right) + 1} \right)}} = \frac{{\left( {k + 1} \right)}}{{2\left( {k + 1} \right) + 1}}\)

Now, adding\(\frac{1}{{\left( {2\left( {k + 1} \right) - 1} \right) \cdot \left( {2\left( {k + 1} \right) + 1} \right)}}\)in both sides of the equation (i) or inductive hypothesis.

\(\begin{array}{c}\left\{ \begin{array}{l}\frac{1}{{1 \cdot 3}} + \frac{1}{{3 \cdot 5}} + ... + \frac{1}{{\left( {2k - 1} \right) \cdot \left( {2k + 1} \right)}}\\ + \frac{1}{{\left( {2\left( {k + 1} \right) - 1} \right) \cdot \left( {2\left( {k + 1} \right) + 1} \right)}}\end{array} \right\} = \frac{k}{{2k + 1}} + \frac{1}{{\left( {2\left( {k + 1} \right) - 1} \right) \cdot \left( {2\left( {k + 1} \right) + 1} \right)}}\\ = \frac{k}{{2k + 1}} + \frac{1}{{\left( {2k + 1} \right) \cdot \left( {2k + 3} \right)}}\\ = \frac{{k(2k + 3) + 1}}{{\left( {2k + 1} \right) \cdot \left( {2k + 3} \right)}}\\ = \frac{{2{k^2} + 3k + 1}}{{\left( {2k + 1} \right) \cdot \left( {2k + 3} \right)}}\end{array}\)

Solve further as:

\(\begin{array}{c}\left\{ \begin{array}{l}\frac{1}{{1 \cdot 3}} + \frac{1}{{3 \cdot 5}} + ... + \frac{1}{{\left( {2k - 1} \right) \cdot \left( {2k + 1} \right)}}\\ + \frac{1}{{\left( {2\left( {k + 1} \right) - 1} \right) \cdot \left( {2\left( {k + 1} \right) + 1} \right)}}\end{array} \right\} = = \frac{{\left( {k + 1} \right)\left( {2k + 1} \right)}}{{\left( {2k + 1} \right) \cdot \left( {2k + 3} \right)}}\\ = \frac{{\left( {k + 1} \right)}}{{\left( {2k + 3} \right)}}\\ = \frac{{\left( {k + 1} \right)}}{{2\left( {k + 1} \right) + 1}}\end{array}\)

From the above, it is clear that\(P\left( {k + 1} \right)\)is also true

Hence,\(P\left( {k + 1} \right)\)is true under the assumption that\(P\left( k \right)\)is true. This completes the inductive step.

Hence, it is proved that\(\frac{1}{{1 \cdot 3}} + \frac{1}{{3 \cdot 5}} + ... + \frac{1}{{\left( {2n - 1} \right) \cdot \left( {2n + 1} \right)}} = \frac{n}{{2n + 1}}\)whenever\(n\)is a positive integer.