Question

Use mathematical induction to show that \[\frac{1}{{1 \cdot 4}} + \frac{1}{{4 \cdot 7}} + ... + \frac{1}{{\left( {3n - 2} \right)\left( {3n + 1} \right)}} = \frac{n}{{3n + 1}}\] whenever nis a positive integer.

Short answer

It is shown that\[\frac{1}{{1 \cdot 4}} + \frac{1}{{4 \cdot 7}} + ... + \frac{1}{{\left( {3n - 2} \right)\left( {3n + 1} \right)}} = \frac{n}{{3n + 1}}\]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

\[\frac{1}{{1 \cdot 4}} + \frac{1}{{4 \cdot 7}} + ... + \frac{1}{{\left( {3n - 2} \right) \cdot \left( {3n + 1} \right)}} = \frac{n}{{3n + 1}}\]

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:

\[\begin{array}{l}\frac{1}{{1 \cdot 4}} = \frac{1}{{3\left( 1 \right) + 1}}\\\frac{1}{4} = \frac{1}{4}\end{array}\]

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.

\(P\left( k \right) \to P\left( {k + 1} \right)\)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\(k\).

\[\frac{1}{{1 \cdot 4}} + \frac{1}{{4 \cdot 7}} + ... + \frac{1}{{\left( {3k - 2} \right) \cdot \left( {3k + 1} \right)}} = \frac{k}{{3k + 1}}\] …. (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.

\[\frac{1}{{1 \cdot 4}} + \frac{1}{{4 \cdot 7}} + ... + \frac{1}{{\left( {3\left( {k + 1} \right) - 2} \right) \cdot \left( {3\left( {k + 1} \right) + 1} \right)}} = \frac{{\left( {k + 1} \right)}}{{3\left( {k + 1} \right) + 1}}\]

Now, adding\[\frac{1}{{\left( {3\left( {k + 1} \right) - 2} \right) \cdot \left( {3\left( {k + 1} \right) + 1} \right)}}\]in both sides of the equation (i) or inductive hypothesis.

\[\begin{array}{c}\left\{ \begin{array}{l}\frac{1}{{1 \cdot 4}} + \frac{1}{{4 \cdot 7}} + ... + \frac{1}{{\left( {3k - 2} \right) \cdot \left( {3k + 1} \right)}}\\ + \frac{1}{{\left( {3\left( {k + 1} \right) - 2} \right) \cdot \left( {3\left( {k + 1} \right) + 1} \right)}}\end{array} \right\} = \frac{k}{{3k + 1}} + \frac{1}{{\left( {3\left( {k + 1} \right) - 2} \right) \cdot \left( {3\left( {k + 1} \right) + 1} \right)}}\\ = \frac{k}{{3k + 1}} + \frac{1}{{\left( {3k + 1} \right) \cdot \left( {3k + 4} \right)}}\\ = \frac{{k\left( {3k + 4} \right) + 1}}{{\left( {3k + 1} \right) \cdot \left( {3k + 4} \right)}}\\ = \frac{{3{k^2} + 4k + 1}}{{\left( {3k + 1} \right) \cdot \left( {3k + 4} \right)}}\end{array}\]

Solve further as:

\[\begin{array}{c}\left\{ \begin{array}{l}\frac{1}{{1 \cdot 4}} + \frac{1}{{4 \cdot 7}} + ... + \frac{1}{{\left( {3k - 2} \right) \cdot \left( {3k + 1} \right)}}\\ + \frac{1}{{\left( {3\left( {k + 1} \right) - 2} \right) \cdot \left( {3\left( {k + 1} \right) + 1} \right)}}\end{array} \right\} = \frac{{\left( {3k + 1} \right)\left( {k + 1} \right)}}{{\left( {3k + 1} \right) \cdot \left( {3k + 4} \right)}}\\ = \frac{{\left( {k + 1} \right)}}{{\left( {3k + 4} \right)}}\\ = \frac{{\left( {k + 1} \right)}}{{3\left( {k + 1} \right) + 1}}\end{array}\]

From the above, we can see 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\[\frac{1}{{1 \cdot 4}} + \frac{1}{{4 \cdot 7}} + ... + \frac{1}{{\left( {3n - 2} \right) \cdot \left( {3n + 1} \right)}} = \frac{n}{{3n + 1}}\]whenever\(n\)is a positive integer.