Question

For which positive integers nis \(n + 6 < {{\left( {{n^2} - 8n} \right)} \mathord{\left/

{\vphantom {{\left( {{n^2} - 8n} \right)} {16}}} \right.

\kern-\nulldelimiterspace} {16}}?\)

Prove your answer using mathematical induction.

Short answer

For\(n \ge 28\),\(n + 6 < {{\left( {{n^2} - 8n} \right)} \mathord{\left/

{\vphantom {{\left( {{n^2} - 8n} \right)} {16}}} \right.

\kern-\nulldelimiterspace} {16}}\)and it is proved using mathematical induction.

Step-by-step solution

Principle of Mathematical Induction

To prove that\(P\left( n \right)\)is true for all positive integers n, where\(P\left( n \right)\)is a propositional function, we complete two steps:

Basis Step:

We verify that\(P\left( 1 \right)\)is true.

Inductive Step:

We show that the conditional statement\(P\left( k \right) \to P\left( {k + 1} \right)\)is true for all positive integers k.

Finding the value of \(n\) 

Given inequality is,

\(\begin{array}{c}n + 6 < \frac{{\left( {{n^2} - 8n} \right)}}{{16}}\\16\left( {n + 6} \right) < {n^2} - 8n\\16n + 96 < {n^2} - 8n\\{n^2} - 24n - 96 > 0\\n > 27.49, - 3\end{array}\)

Since we need a positive integer value of\(n\). Therefore\(n \ge 28\)

Hence for\(n \ge 28\),\(n + 6 < {{\left( {{n^2} - 8n} \right)} \mathord{\left/

{\vphantom {{\left( {{n^2} - 8n} \right)} {16}}} \right.

\kern-\nulldelimiterspace} {16}}\)

Proving the basis step

Given statement is

\(n + 6 < \frac{{\left( {{n^2} - 8n} \right)}}{{16}}\)

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

Since integer\(n \ge 28\)

Therefore, in the basis step, we need to prove that\(P\left( {28} \right)\)is true

For finding statement\(P\left( {28} \right)\)substituting\(28\)for\(n\)in the statement

Therefore, the statement\(P\left( {28} \right)\)is

\(\begin{array}{l}28 + 6 < \frac{{\left( {{{28}^2} - 8\left( {28} \right)} \right)}}{{16}}\\34 < \frac{{\left( {784 - 224} \right)}}{{16}}\\34 < \frac{{560}}{{16}}\\34 < 35\end{array}\)

The statement \(P\left( {28} \right)\) is true this is also known as the basis step of the proof.

Proving 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

\(k + 6 < \frac{{\left( {{k^2} - 8k} \right)}}{{16}}\)

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

Therefore, replacing\(k\)with\(k + 1\)in the left-hand side of the statement

\(\begin{array}{c}\left( {k + 1} \right) + 6 = \left( {k + 6} \right) + 1\\ < \frac{{\left( {{k^2} - 8k} \right)}}{{16}} + 1\\ < \frac{{\left( {{k^2} - 8k} \right) + 16}}{{16}}\\ < \frac{{{k^2} + 2k + 1 - 10k + 15}}{{16}}\\ < \frac{{\left( {k + 1} \right) - 10k + 15}}{{16}}\end{array}\)

Since \( - 10k + 15 < - 8k - 8\) for\(k \ge 28\)

\(\begin{array}{c}\left( {k + 1} \right) + 6 < \frac{{\left( {k + 1} \right) - 8k - 8}}{{16}}\\\left( {k + 1} \right) + 6 < \frac{{\left( {k + 1} \right) - 8\left( {k + 1} \right)}}{{16}}\end{array}\)

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 shown that\(n + 6 < \frac{{\left( {{n^2} - 8n} \right)}}{{16}}\)for\(n \ge 28\)