- #1
rad0786
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Okay, so we are covering proof by induction, and i need some ones help on it covering inequalities.
(a) (2^n) ≤ n! , n≥4
Base Step: sub in n=1 and yes, it works!
Inductinve step: assume (2^n) ≤ n! and show (2^(k+1)) ≤ (k+1)! ,K≥4 holds.
(2^(k+1)) ≤ (k+1)!
(2)(2^k) ≤ (K!)(K+1)
So the bold part is the original equation. In the above, the left side was multiplied by 2 and the right side was multiplied by (K+1). So, anything multiplied by 2 is less than anything multiplied by (K+1), K≥4. Is this proof reasonable?
---------------------------
I initially tried this:
(2^(k+1))
= 2(2^K)
= ?
= ?
≤ (k+1)!
What I am trying to show is to work my way from (2^(k+1)) by expanding it and eventually show that it is less that (K+1)!
somebody care to comment?
Thanks
(a) (2^n) ≤ n! , n≥4
Base Step: sub in n=1 and yes, it works!
Inductinve step: assume (2^n) ≤ n! and show (2^(k+1)) ≤ (k+1)! ,K≥4 holds.
(2^(k+1)) ≤ (k+1)!
(2)(2^k) ≤ (K!)(K+1)
So the bold part is the original equation. In the above, the left side was multiplied by 2 and the right side was multiplied by (K+1). So, anything multiplied by 2 is less than anything multiplied by (K+1), K≥4. Is this proof reasonable?
---------------------------
I initially tried this:
(2^(k+1))
= 2(2^K)
= ?
= ?
≤ (k+1)!
What I am trying to show is to work my way from (2^(k+1)) by expanding it and eventually show that it is less that (K+1)!
somebody care to comment?
Thanks