[Bleak Lemming]

Just can't get into proof by mathematical induction

MS221 2009 Q16 c

Prove 2^n < n! for n greater or equal to 5 (4 marks)

First solve for p(5) done & true,

So presume p(k) true

For for p(k+1) I would say

2^k+1 = (k+1)!
thus
2x2^k = k!(k+1)

then not sure where to go.. think i'd get 2 out of the 4 marks at least ...

[Allofthem]

The idea is to extend what we'd do in a particular case,
for example, 2^5 < 5!, so 2*2^5< 5! * 2 < 5! * 6, 2^6 < 6!

For induction we write a general argument.

Edit: And of course show that it's true for some case, important!
As a side note, why was there all those nonsense words we were to write at A level? It was a good 3 lines of needless rubbish, not seen it at uni, reminds me of mixed numbers.

[BabyMaths]

but

so

[matt2k8]

Using your notation, I'd write something like:

We know the statement P(5) is true. Now for k >= 5, suppose P(k) is true. We WTS P(k+1) must also be true.

2^k+1 = 2 * 2^k < 2 * k! as P(k) is true
< (k+1) * k! as 2 < k
= (k+1)!

so P(k+1) is true. So we have P(5) true, and for k >= 5, P(k) true implies P(k+1) true; hence by induction, P(n) is true for all n >= 5.