Recent content by MP14

Yes! Thank you! (2(k+1))! = (2k+2)! = (2k+2)(2k+1)! = (2k+2)(2k+1)(2k)! = (4k^2 + 6k + 2)(2k)! < (4k^2 + 8k + 4)(2k)! < (4k^2 + 8k + 4)(2^(2k))(k!)^2 (based on assumption) and the rest is algebra! Thanks again!

Yes my mistake, A = (2k+2)(2k+1) = (4k^2 + 6k + 2) and B = (4k^2 + 8k + 4), so B is larger.

Solving for A I get A = 1 + 1/k, and for B I get B = 4(k+1)^2. I don't think that we can multiply the inequality by k, so is there another way to approach this?

I'm not too sure... I tried multiplying both sides by 4, because if you do that you would end up getting: 4(2k)! < 4(2^(2k))*(k!)^2 4(2k)! < (2^2)*(2^(2k))*(k!)^2 4(2k)! < (2^(2k+2))*(k!)^2 4(2k)! < (2^(2(k+1)))*(k!)^2 But then that still leaves me with (2k) on the left side and (k) on the...

Prove by mathematical induction: (2n)! < (2^(2n))*(n!)^2 , for all n=2,3,4... I know that to start you must prove that it is true for n=2, (2*2)! = 24 < 64 = (2^4)(2!)^2 Then you assume that n=k and show tha n=k implies that n=(k+1) (2k)! < (2^(2k))*(k!)^2 ... At this point I...