How can I prove that
For all $n \in \Bbb N$ such that $10<n$ then $0 < n^2 -13n +24$ by induction.
This is what I have so far:
Base Case:
$n = 11$
If $n = 11$ then:
- $0 < n^2 -13n +24$
- $0 < (11)^2 - 13 (11) + 24$
- $0 < 2$
Induction Hypothesis:
$0 < k^2 -13k +24$
Inductive Step:
$0 < (k+1)^2 -13(k+1) +24$
I’m supposed to get to this $2k-12$, but I don’t understand how, and after that I think we need to do something else.
For the base case $n=11$, $f=121-143+24=1$ so the base case is positive.
Then for $n+1$, we have $f(n+1)=f(n)+2n-14$
Well, clearly $2n>14$ for all $n>11$ so $f$ is strictly increasing, QED.