prove by Induction that n! ≥ 2^(n-1) n ≥ 1
Oct 4, 2013 #3 kudzie adore 2 0 I managed to prove for n= 1 and is true for n=k k!≥ 2^ (k-1) and I assumed that n=k to be true then for n= k+1 its (k+1)! ≥ 2^[(k+1)-1] proof for n=(k+1) (k!)(k+1) ≥ _____? the problem is that how do we reach the proof for (k+1)
I managed to prove for n= 1 and is true for n=k k!≥ 2^ (k-1) and I assumed that n=k to be true then for n= k+1 its (k+1)! ≥ 2^[(k+1)-1] proof for n=(k+1) (k!)(k+1) ≥ _____? the problem is that how do we reach the proof for (k+1)
Oct 4, 2013 #4 tiny-tim Science Advisor Homework Helper 25,836 255 hi kudzie adore! welcome to pf! (try using the X^{2} button just above the Reply box ) hint: if an equation is true, then multiplying both sides by the same factor will still be true
hi kudzie adore! welcome to pf! (try using the X^{2} button just above the Reply box ) hint: if an equation is true, then multiplying both sides by the same factor will still be true