Induction q: n! >or= 2^(n-1) (1 Viewer)
- Thread starter enigma_1
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Squar3root
realest nigga
not sure if this is correct but
do all the assume stuff and n=1 and all that
now (k+1)!=k!(k+1)
greater than or equal to (k+1)*2^(k-1)
expand above we get k*2^(k-1) + 2^(k-1)
which is greater than or equal to 2^(k-1)
as required
maybe Sy123 can correct me
do all the assume stuff and n=1 and all that
now (k+1)!=k!(k+1)
greater than or equal to (k+1)*2^(k-1)
expand above we get k*2^(k-1) + 2^(k-1)
which is greater than or equal to 2^(k-1)
as required
maybe Sy123 can correct me
enigma_1
~~~~ Miss Cricket ~~~~
Cheers mate, but what is this step?
And is this a common induction question that could rock up in half yearlies? If so I'm dead. I don't get these types of inductions, I like the normal ones haha
integral95
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For n = 1 LHS = 1 and RHS = 1 so n= 1 is true
Now Assume true for n = k
Prove true for n = k+1, that is!\geq 2^{k} )
! \\ = (k+1)(k)! \\ \geq (k+1)2^{k-1} \ \ \ \ \ \ \ (From assumption)\\ \geq 2.2^{k-1} \ \ \ \ \ \ \ \ \ (k+1\geq 2)\\ = 2^k )
Therefore true for n = k+1
I wouldn't say it's a common maths induction question, but it's not rare either
Now Assume true for n = k
Prove true for n = k+1, that is
Therefore true for n = k+1
I wouldn't say it's a common maths induction question, but it's not rare either
Last edited:
Squar3root
realest nigga
Yeah the answer above is more correct, the part in red is bringing in the assumption. I forgot to add one to the power - my bad
Squar3root
realest nigga
Yes, I have seen it in terry lee, Cambridge and fitzpatrick and iirc a past hsc paper