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HSC 2013-14 MX1 Marathon (archive) (3 Viewers)

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Sy123

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Re: HSC 2013 3U Marathon Thread



 

RealiseNothing

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Re: HSC 2013 3U Marathon Thread

Considering Pascal's triangle, we know that the sum of two adjacent numbers is the number directly below them. Hence we can write the question as the sum of the numbers in the row above it:



This is just the sum of the row of Pascal's triangle, which is
 

Sy123

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Re: HSC 2013 3U Marathon Thread

Considering Pascal's triangle, we know that the sum of two adjacent numbers is the number directly below them. Hence we can write the question as the sum of the numbers in the row above it:



This is just the sum of the row of Pascal's triangle, which is
Very elegant, nice work, my way was a little longer by considering expansions of (1+x)^2n and (1-x)^2n
 
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Sy123

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Sy123

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Re: HSC 2013 3U Marathon Thread

A nice result I've proven



 

Sy123

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Re: HSC 2013 3U Marathon Thread

Err, not true. The last term on the LHS alone is greater than the RHS...
I considered the geometric series:



Equate co-efficient of x^n on both sides of equation. Anything illegal in the proof?
 

seanieg89

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Re: HSC 2013 3U Marathon Thread

Yes. Check the number of terms on the LHS, there are n+1, not n. A minor change needs to be made to your formula for the sum of a finite geometric series.
 

Sy123

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Re: HSC 2013 3U Marathon Thread

Yes. Check the number of terms on the LHS, there are n+1, not n. A minor change needs to be made to your formula for the sum of a finite geometric series.
:/ Can't believe I missed that, thanks for the heads up. I'll ask this question again once everyone has forgotten about it.

======================

New result:



 

Sy123

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Re: HSC 2013 3U Marathon Thread



(The signs alternate in pairs)
(Provide proof of your value, either with latex or a brief outline of steps taken)
 
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Sy123

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Re: HSC 2013 3U Marathon Thread

:/ Can't believe I missed that, thanks for the heads up. I'll ask this question again once everyone has forgotten about it.

======================

New result:



Note that the LHS is:

(which we can derive easily)

Use u-substitution, we yield the RHS.
 

RealiseNothing

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Re: HSC 2013 3U Marathon Thread



(The signs alternate in pairs)
(Provide proof of your value, either with latex or a brief outline of steps taken)
Consider them in groups of four. Now consider the general case:









Now we will note that:



So the sum of each group of four is just two times the sum of that group. Hence if we were to do it to all groups of four up to 100, we would get two times the sum of all the numbers from 1 to 100:



This is just double the 100th triangular number, which is just

So the answer is
 

Sy123

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Re: HSC 2013 3U Marathon Thread

Consider them in groups of four. Now consider the general case:









Now we will note that:



So the sum of each group of four is just two times the sum of that group. Hence if we were to do it to all groups of four up to 100, we would get two times the sum of all the numbers from 1 to 100:



This is just double the 100th triangular number, which is just

So the answer is
Elegant, nice work

Alternatively:



Difference of 2 squares



Note that first bracket in each term is 2, take common:

 

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Capt Rifle

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Re: HSC 2013 3U Marathon Thread

<a href="http://www.codecogs.com/eqnedit.php?latex=Find \int cos^42x\, sin^32x\, \, dx\, \: using\, the\, subsitution\, u=cos2x" target="_blank"><img src="http://latex.codecogs.com/gif.latex?Find \int cos^42x\, sin^32x\, \, dx\, \: using\, the\, subsitution\, u=cos2x" title="Find \int cos^42x\, sin^32x\, \, dx\, \: using\, the\, subsitution\, u=cos2x" /></a>

Edit: This is at a 3U difficulty, right guys?
 
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