Question 1183367
.
prove that n^2 ≤ n! for all n ≥ 4 using mathematical induction
~~~~~~~~~~~~~~~



<U>Proof</U>  &nbsp;&nbsp;&nbsp;&nbsp;(from  https://people.cs.umass.edu/~barring/cs240/induction_sol.pdf )


<pre>
Note first that:


  if n = 0, then 02 = 0 and 0! = 1.
  if n = 1, then 12 = 1 and 1! = 1.
  if n = 2, then 22 = 4 and 2! = 2.
  if n = 3, then 32 = 9 and 3! = 6.


We prove by induction on n that  {{{n^2}}} ≤ n!  for all  n ≥ 4.


<U>Basis step</U> : 

    {{{4^2}}} = 16 and 4! = 24


<U>Inductive hypothesis</U> : 

    Assume for some integer k ≥ 4 that  {{{k^2}}} ≤ k!


<U>Inductive step</U> :


       (k + 1)! = (k + 1)k! 

     ≥ {{{(k + 1)k^2}}}

     = {{{k^2*k + k^2}}}

     ≥ {{{4^2*k + k^2}}}

     = {{{15k + k + k2}}}

     ≥ {{{15k + 1 + k^2}}}

     ≥ {{{2k + 1 + k^2}}}

     = {{{(k + 1)^2}}}.


According to the method of Mathematical induction, the proof is completed.
</pre>

Solved.