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Re: HSC 2013 4U Marathon Haha, I will post solutions maybe 1-2 nights after I post them then. So I will do the one in the 3U thread now

Re: HSC 2013 4U Marathon If there is enough demand for it sure, definitely :)

Re: HSC 2013 4U Marathon $A sequence is defined recursively as follows$ F_0 = 0 \ \ F_1=1 F_n=F_{n-1}+F_{n-2} \ \ \ n \geq 2 $i) Write down the first 7 numbers of this sequence$ \ \ \ \fbox{1} $ii) Consider a power series in$ \ z \ $whereby the co-efficients are the numbers of the...

Re: HSC 2013 4U Marathon Are you sure? Wolfram Alpha says it is correct

Re: HSC 2013 3U Marathon Thread For that one: Using that identity, equate the co-efficient of x^2 on both sides: Now to select 2 x's from the RHS, we need to either pick x from 1 (1+x)^n and 1 x from (1+x)^n, this makes x^2 And the ways we can do this is: First pick 2 brackets from the n...

Re: HSC 2013 4U Marathon Relax I only got to have time to think of your problem now, its actually very simple: a \leq 1 a+b \leq 5 a+b+c \leq 14 a+b+c+d \leq 30 Now, using subtraction, it follows that: a\leq 1 \ \ b \leq 4 \ \ c \leq 9 \ \ d \leq 16 Conveniently \sqrt{a} \leq 1 \ \...

Re: HSC 2013 4U Marathon Are you sure that isn't the answer?

Re: HSC 2013 4U Marathon $i) Show that$ 1+2z+3z^2+\dots+nz^{n-1} = \frac{nz^{n+1}-(n+1)z^n +1}{(1-z)^2} \ \ \ |z|<1 \ \ \ \fbox{2} $ii) Hence find in terms of $ \ z \ \ \ \ \sum_{k=1}^{\infty} kz^k \ \ \ \fbox{1} $iii) The Triangular numbers are a sequence of numbers defined by the...

Re: HSC 2013 4U Marathon I want to do this but I don't want to let the marathon die lol (even though its on the verge of death at the moment) ================ $Find rational$ \ \ x \ \ $such that$ \sqrt{x} = \sqrt{2+\sqrt{3}} + \sqrt{2-\sqrt{3}}

Re: HSC 2013 3U Marathon Thread I'm assuming in your last line its a typo for (as fixed above) the last term on the LHS of your last term used to be \binom{n}{n-m-1} So I'm assuming its: \binom{n-1}{n-m-1} Its an interesting method, but I will leave it to someone else to say whether its...

Re: HSC 2013 4U Marathon I have already completed the 3U course which is why I am able to do most of the Harder 3U questions on here, but in terms of the 4U syllabus I only know what I have been taught in class + Integration

Re: HSC 2013 3U Marathon Thread I'm not a fan of it ============================ $Prove using the Binomial Theorem/Combinatorics that$ \sum_{j=m}^{n} \binom{j}{m} \equiv \binom{m}{m} + \binom{m+1}{m} + \binom{m+2}{m}+ \dots + \binom{n}{m} = \binom{n+1}{m+1}

Re: HSC 2013 4U Marathon YesI should of put that in the question, good work anyway. I wil leave the Conics question to someone else who has learnt all of it

Re: HSC 2013 4U Marathon $Prove that$ \ \ \frac{a+b}{2} \geq \sqrt{ab} $Using the above result, hence prove that$ \frac{a+b+c+d}{4} \geq \sqrt[4]{abcd}

Re: HSC 2013 4U Marathon Go right ahead

Re: HSC 2013 4U Marathon See how you can bold text using the [.B] amd [./B] (without the dots), like this: bold (to see the tags quote this post) So for latex, use [.tex] and [./tex] (without the dots), like this: $THIS IS LATEX$ \ \ \frac{\pi^2}{6} There is a short guide on how to use it...

Re: HSC 2013 4U Marathon For the second one, consider what you did for the RHS of your expression, but do that for I instead

Re: HSC 2013 4U Marathon An easier one $Find all solutions to$ (1+z)^n - (1-z)^n = 0

Re: HSC 2013 3U Marathon Thread I never implied otherwise. ?

Are you sure they will drop you? You need to pick it up in the HYE to convince them that you can do 2U maths, make sure to improve your algebra skills and interpretation, because that is essentially all of what 2U maths really is, there aren't any tricks or clever things you need to find, its...