evil1
Dec7-03, 05:47 AM
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
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