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Think I have parts 1 and 2, struggling with the induction :S Any help greatly appreciated!

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- Sep 20th 2013, 11:05 AMTheManFromScotCombinatorics Induction Proof
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Think I have parts 1 and 2, struggling with the induction :S Any help greatly appreciated! - Sep 20th 2013, 11:41 AMPlatoRe: Combinatorics Induction Proof
- Sep 20th 2013, 12:12 PMTheManFromScotRe: Combinatorics Induction Proof
I am guessing so.. I really can't fathom it out.. When n=1, An = 2 which isn't greater than 4 is as far as I got

- Sep 20th 2013, 12:16 PMPlatoRe: Combinatorics Induction Proof
I think that the base case should be

Have a look at this webpage. - Sep 20th 2013, 12:42 PMFelixFelicis28Re: Combinatorics Induction Proof
There is no need for induction on this question when you can prove it simply by re-writing the inequality but very well...

I am going to assume there was a misprint and they omitted that the result holds true for .

OP, from your induction hypothesis, you have for and you wish to show with the assumption that .

Now, recall that .

Multiply the inequality through by and the result falls out fairly quickly with some manipulation.

@Plato, I think you mean for as equality holds for . - Sep 21st 2013, 05:41 AMTheManFromScotRe: Combinatorics Induction Proof
Attachment 29221

Thanks for the help - where am I going wrong? - Sep 21st 2013, 09:30 AMFelixFelicis28Re: Combinatorics Induction Proof
No problem. There's nothing wrong with your working, you just need to do a bit of deduction for the last bit.

Clearly for

Also, this may be pedantic but I think you could make it a bit clearer that what you've done is that you've**taken**your inductive assumption and you've multiplied that through by and then added 1. - Sep 22nd 2013, 02:54 AMTheManFromScotRe: Combinatorics Induction Proof
Thank you very much!

I;ve had a look at part (d) and tried a sort of reverse induction. I really think I am overly complicating it though :/ - Sep 22nd 2013, 05:31 AMFelixFelicis28Re: Combinatorics Induction Proof
- Sep 22nd 2013, 06:12 AMTheManFromScotRe: Combinatorics Induction Proof
Though I am following the logic, these are jumps I'd never considered.. Now I can;t even manipulate haha! This is quite clearly beyond me, though thank you very much for the input

- Sep 22nd 2013, 06:18 AMFelixFelicis28Re: Combinatorics Induction Proof
It's not beyond you, just keep going at it! Practice makes pefect. ;)

Clearly if .

Substitute for and then you have the rightmost inequality in my above post. You will be able to simplify that down into a fraction, call it . The denominator of will clearly be and so clearly iff the numerator of . Try it, you'll be able to sneak in the result from part C. ;)

Also, I see you're a fellow Scot! Are you studying Mathematics @ Glasgow? - Sep 23rd 2013, 07:42 AMTheManFromScotRe: Combinatorics Induction Proof
Thank you very much :D think I have this one sorted now! I'm not living in Glasgow now, have travelled over the border - can never lose my Scottish roots though, haha

- Sep 25th 2013, 10:34 AMTheManFromScotRe: Combinatorics Induction Proof
Turns out my question 2b falls apart :/

Can anyone shed some light on this?

Thank you - Sep 25th 2013, 10:49 PMFelixFelicis28Re: Combinatorics Induction Proof
- Sep 25th 2013, 11:33 PMTheManFromScotRe: Combinatorics Induction Proof
Hi again :D Thank you very much for the reply, but I managed to readjust my previous method - this involved accounting for the middle term when 'n' is odd, and readjusting - your method is much easier on the eye