1.) Show that n! > or = 3^n for n > or =7
2.) verify for all n > or = 1 the sum of the squares of the first 2n postive interger is given by the formular 1^2 + 2^2 + 3^2 + ....+(2n)^2 = n(2n + 1)(4n + 1) / 3
check n=7
7! >= 3^7
5040 >= 2187✅
If true for n=k, check for n=k+1 (k >= 7)
(k+1)! = k! * (k+1) >= 3^k (k+1) >= 3^k * 3 = 3^(k+1)
why don't you try the other one. Post your work if you get stuck.