Can anyone help me confirm if I've got this one correct? Particularly the part at
A, where I've defined (k + 2)! = (k + 1)!(k + 2).
Many thanks.
Q. (n + 1)!
> 2
^{n}, for n
Attempt: Step 1: For n = 1...
(1 + 1)! = 2! = 2 & 2
^{1} = 2
Since 2
> 2, the statement is true for n = 1.
Step2: Assume the statement is true for n = k, i.e. assume (k + 1)!
> 2
^{k}.
We must now show that the statement is true for n = k + 1,
i.e. ((k + 1) + 1)!
> 2
^{k+1} if (k + 2)!
> 2
^{k+1} A: (k + 2)! = (k + 1)!(k + 2)
> (k + 1)2
^{k}...((k + 1)!
> 2
^{k}...assumed)
If (k + 1)2
^{k} > 2
^{k+1}, then (k + 2)!
> 2
^{k+1}
if (k + 1)2
^{k} > 2
^{k+1}
if (k + 1)2
^{k} - 2
^{k+1} > 0
if 2
^{k}[(k + 1) - 2]
> 0...which is true for k
> 1
=> (k + 2)!
> 2
^{k+1}
Therefore, the statement is true for n = k + 1, if true for n = k. Thus, the statement is true for all n
> 2, n