Prove pa Using Mathematical Induction

[evil1]

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 some k > 10. 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 ……(b) because

> ..?.... because





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

thanks

 

[KLscilevothma]

https://www.physicsforums.com/showthread.php?s=&postid=109155#post109155

 

[tacman]

To complete the proof, we need to show that (k-2) < (k^2 - 3k)/12 + (2k-2)/12. We can rewrite (b) as (k-2) < (k^2 - 3k + 2k - 2)/12, which simplifies to (k-2) < (k^2 - k - 2)/12.

Next, we can factor the numerator of the right side to get (k-2) < (k-2)(k+1)/12. Since k > 10, we know that k+1 > 11, so we can divide both sides by (k+1) to get (k-2)/(k+1) < (k-2)(k+1)/(12(k+1)).

Finally, we can simplify the right side to get (k-2)/(k+1) < (k-2)/12. Since we assumed that k > 10, we know that (k-2)/12 < 1. Therefore, (k-2)/(k+1) < (k-2)/12 < 1, which proves that pa(k+1) is true.

By the principle of mathematical induction, we can conclude that pa is true for all integers n greater than 10, as desired.