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HSC 2013 MX2 Marathon (archive) (3 Viewers)

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RealiseNothing

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Re: HSC 2013 4U Marathon

The way I did it was to use:



Then equate imaginary parts to get the result:



Which is wrong, so idk.
 

Carrotsticks

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Re: HSC 2013 4U Marathon



by De Moivre's Theorem



Sum each one individually (a neat pattern arises). Then add them together and equate imaginary components.
Yep, so quite like the first method I mentioned. That method SHOULD work, but there is much larger room for errors to be made. Try finding a closed form again, but using the second method I provided.
 

RealiseNothing

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Re: HSC 2013 4U Marathon

by a telescoping sum.

Now expanding the binomial first instead of using a telescoping sum gives:



Now if we apply the summation, we sum each one individually, that is:



Now using our definition of we obtain:



Using the symmetry property of binomial expansions gives:



Now since and we get:



Now consider the second part of the equation:

by definition.

Hence if we add the two parts together we obtain:



So the two terms on the end cancel out and give:



But this was equal to the value of the telescoping sum, so we can equate the two and get:



Taking everything from the RHS to the LHS except for the last term gives us the required result of:



Which can be re-written in the form asked:



Quod Erat Demonstrandum.
 
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RealiseNothing

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Re: HSC 2013 4U Marathon

I'll do part (ii) when my hands can be bothered latexing again lol.
 

RealiseNothing

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Re: HSC 2013 4U Marathon

How do I make it so that it is all on one line and doesn't go to the next?
 

Carrotsticks

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Re: HSC 2013 4U Marathon

Don't think you can do that. It auto formats and goes into the next line when too long.
 

RealiseNothing

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Re: HSC 2013 4U Marathon

Given:



Integrating the RHS gives:





Therefore:



Thus we can integrate the following to obtain on the LHS:



Since the RHS is a polynomial in 'n' itself with highest degree 'k', it follows that after integration, the highest power will become 'k+1', and hence is a polynomial of degree 'k+1' in 'n'.
 

bleakarcher

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Re: HSC 2013 4U Marathon

Given:



Integrating the RHS gives:





Therefore:



Thus we can integrate the following to obtain on the LHS:



Since the RHS is a polynomial in 'n' itself with highest degree 'k', it follows that after integration, the highest power will become 'k+1', and hence is a polynomial of degree 'k+1' in 'n'.
You could have also replaced k with k+1.
 

Sy123

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Re: HSC 2013 4U Marathon

Given:



Integrating the RHS gives:





Therefore:



Thus we can integrate the following to obtain on the LHS:



Since the RHS is a polynomial in 'n' itself with highest degree 'k', it follows that after integration, the highest power will become 'k+1', and hence is a polynomial of degree 'k+1' in 'n'.
Nice work! Your approach to the first and second one was different to mine, I actually ripped this question off of STEP 2008 III Q2
 

Carrotsticks

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Re: HSC 2013 4U Marathon

yeah, whichever ones you have to spare. I don't mind if it's a 2012 one.
Not bothered atm to dig up the old BOS Trials thread, and can't find my spare question file for 2013 actually.
 

Sy123

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Re: HSC 2013 4U Marathon

I made this question just now:





 
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