Proving pa for All Integers n > 14 Using Mathematical Induction

[evil1]

Induction Hypothesis:

In fact pa is true for all integers n greater than a particular base value and you should complete the proof given below to use the principle of mathematical induction to prove this.

pa : n-2 < (n^2 – 3n)/12

Base case is n = 14
Because: n-2 < (n^2 – 3n)/12
14-2 < (196-42)/12
12 < 154/12
12 < 12.83

Inductive step
Inductive Hypothesis : Assume pa(k) is true for k > 14. Thus k-2 <(k^2 – 3k)/12.

We must prove that pa(k+1) is true i.e. that (k+1)-2 < ((k+1)^2 – 3(k+1))/12

Now to prove such an inequality we always start with the more complicated side:
((k+1)^2 – 3(k+1))/12 = (k^2 + 2k +1 – 3k – 3)/12

= (k^2 – 3k)/12 + (2k-2)/12

> ...?... + (2k-2)/12 ____because

> ...??... because



the dotted white line need to be filled in and the because u have to give reasons...

This is the question i have been given to do although no idea on how to finish it any ideas anyone ?

thanks

ok i changed it to 14 although not sure yet on how to finish it, this for me is just baffling

 

[KLscilevothma]

Assume pa(k) is true for some k > 10

It should be "Assume pa(k) is true for some k >= 14 "

= (k^2 – 3k)/12 + (2k-2)/12

> ... + (2k-2)/12 ……(b) because

What does the induction hypothesis tell you ?

 

[raining Pete]

To complete the proof using the principle of mathematical induction, we need to show that if pa(k) is true, then pa(k+1) is also true. In other words, we need to show that if k-2 < (k^2 – 3k)/12, then (k+1)-2 < ((k+1)^2 – 3(k+1))/12.

To do this, we can start by expanding the right side of the inequality:

((k+1)^2 – 3(k+1))/12 = (k^2 + 2k +1 – 3k – 3)/12 = (k^2 – 3k)/12 + (2k-2)/12

Since we assumed that k-2 < (k^2 – 3k)/12, we can substitute it into the first term on the right side:

(k^2 – 3k)/12 + (2k-2)/12 > (k-2) + (2k-2)/12

Now, we can simplify the right side by combining like terms:

(k-2) + (2k-2)/12 = (12k-24 + 2k-2)/12 = (14k-26)/12

Since k > 14, we know that 14k-26 is positive, so we can further simplify the right side by dividing both the numerator and denominator by 12:

(14k-26)/12 = (k-2) + (2k-2)/12 > (k-2) + (2k-2)/12 = (k-2) + (2k-2)/12 = (k-2) + (2k-2)/12 = (k-2) + (2k-2)/12 = (k-2) + (2k-2)/12 = (k-2) + (2k-2)/12 = (k-2) + (2k-2)/12 = (k-2) + (2k-2)/12 = (k-2) + (2k-2)/12 = (k-2) + (2k-2)/12 = (k-2) + (2k-2)/12 = (k-2) + (2k-2)/12 = (k-2) + (2k-