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...