Prove using the principle of mathematical induction: (i) The number of diagonals of a convex polygon with n vertices is n(n − 3)/2, for n ≥ 4, (ii) 2n < n! for for all n > k > 0, discover the value of k before doing induction.

Step-by-step explanation:Proof for i)We will prove by mathematical induction that, for every natural [tex]n\geq 4[/tex], the number of diagonals of a convex polygon with n vertices is [tex]\frac{n(n-3)}{2}[/tex].In this proof we will use the expression d(n) to denote the number of diagonals of a convex polygon with n vertices Base case:First, observe that:, for n=4, the number of diagonals is [tex]2=\frac{n(n-3)}{2}[/tex]Inductive hypothesis:  Given a natural [tex]n \geq 4[/tex],[tex]d(n)=\frac{n(n-3)}{2}[/tex]Now, we will assume the induction hypothesis and then use this assumption, involving n, to prove the statement for n + 1.Inductive step:Observe that, given a convex polygon with n vertices, wich we will denote by P(n), if we add a new vertix (transforming P(n) into a convex polygon with n+1 vertices, wich we will denote by P(n+1)) we have that:Every diagonal in P(n) will still be a diagonal in P(n+1). One (and only one) side of P(n) will be a diagonal in P(n+1).There would be an extra n-2 diagonals (those that connect with the new added vertix).Because of these observation we know that, for every [tex]n\geq 4[/tex],[tex]d(n+1)=d(n)+1+(n-2)=d(n)+n-1[/tex]Therefore:[tex]d(n+1)=d(n)+n-1=\frac{n(n-3)}{2}+n-1=\frac{n^2-3n+2n-2}{2}=\frac{n^2-n-2}{2}=\frac{(n+1)(n-2)}{2}[/tex]With this we have proved our statement to be true for n+1.    In conlusion, for every natural [tex]n \geq 4[/tex],[tex]d(n)=\frac{n(n-3)}{2}[/tex]Proof for ii)Observe that:For n=1 [tex]2n=2>1=n![/tex]For n=2 [tex]2n=4>2=n![/tex]For n=3 [tex]2n=6=n![/tex]Then, the statement is not true for n=1,2,3.We will prove by mathematical induction that, for every natural [tex]n \geq 4[/tex], [tex]2n<n![/tex].Base case:For n=4, [tex]2n=8<24=n![/tex]Inductive hypothesis:  Given a natural [tex]n \geq 4[/tex], [tex]2n<n![/tex]Now, we will assume the induction hypothesis and then use this assumption, involving n, to prove the statement for n + 1.Inductive step:Observe that, [tex]n!+2\leq (n+1)! \iff n!+2\leq n!(n+1) \iff 1+\frac{2}{n!}\leq n+1 \iff 2\leq n*n![/tex]wich is true as we are assuming [tex]n\geq 4[/tex]. Therefore:[tex]2(n+1)=2n+2<n!+2\leq (n+1)![/tex]With this we have proved our statement to be true for n+1.    In conlusion, for every natural [tex]n \geq 4[/tex],[tex]2n<n![/tex]