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Re: HSC 2013 4U Marathon To be fair, I'll put this first part: $A series$ \ a_1, a_2, a_3, \dots \ \ $diverges if$ \ \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right | > 1 $Hence or otherwise$

Re: HSC 2013 3U Marathon Thread I'm pretty sure you get (-1)^n \binom{2n}{n} The n+1th term in that binomial expansion in [3] is \binom{2n}{n} (-1)^n (x^2)^n

Re: HSC 2013 3U Marathon Thread When integrating (1+x)^n, we get \frac{(1+x)^{n+1}}{n+1} + c = x \times ($Something$) Then x=0 to find c= - 1/(n+1) Then, \frac{(1+x)^n - 1}{n+1} = \frac{n}{0}x + \frac{1}{2} \binom{n}{1} x^2 + \dots + \binom{n}{n} \frac{1}{n+1} EDIT: Yep I'm wrong I...

Re: HSC 2013 3U Marathon Thread Yep the steps are correct, however when you do the (1) - (2) we get 2 times the sum on the LHS = 2^n/ (n+1) dividing by 2 gives 2^(n-1) ==== $Prove$ \binom{2n}{0}^2 - \binom{2n}{1}^2 + \binom{2n}{2}^2 - \binom{2n}{3}^2 + \dots - \binom{2n}{2n-1}^2 +...

Re: HSC 2013 4U Marathon Yeah this was it, I was able to show that u_{n+1} > u_n and u_n > 0 , is this enough to justify convergence given the of the final step in this problem? (which I don't want to type out here just yet) I guess this is probably a form of monotone convegence

Re: HSC 2013 4U Marathon A good Sydney grammar question u_{n+1} = u_n + u_n^2, \ u_1 = \frac{1}{3} $Find$ \ \sum_{n=1}^{\infty} \frac{1}{1+u_n}

Re: HSC 2013 3U Marathon Thread $Prove$ \ \binom{n}{0} + \frac{1}{3} \binom{n}{2} + \frac{1}{5} \binom{n}{4} + \frac{1}{7} \binom{n}{6} + \dots = \frac{2^{n-1}}{n+1}

Re: HSC 2013 4U Marathon $Using the result or otherwise$ \frac{a_1 + a_2 + \dots + a_n}{n} \geq \sqrt[n]{a_1 a_2 a_3 \dots a_n} \ \ $for positive$ \ a_k \ k=1,2, \dots , n $Prove that for all positive integers$ \ n n^n \geq (n+1)^{n-1} $for what value of$ \ n \ $does equality hold? $

That is true, I hope my comments have not come off as insensitive, I really do appreciate the work that teachers put in. However compared to other commercial trials and previous CSSA papers, this year's paper simply does not cut it. And I do concede that perhaps its a little unfair to compare...

Re: HSC 2013 3U Marathon Thread I don't think so, I could be wrong but I'm getting what the question sasys.

Even James Ruse, they may have the odd good question, but overall they aren't very good (which is why schools buy trial papers, but CSSA didn't do as well with the trials this year)

bump

Only one I've seen is Sydney Grammar (though the 2012 one is pretty bad) Oh and Moriah but I've only seen their 2001 paper which was great (or was that the 2003 one im not sure), I haven't seen their later papers though

No idea, I didn't mean to imply that, rather that the trials from most selective schools whose trials are available on Bos are nearly all of low quality with a select few being ok.

Re: HSC 2013 4U Marathon I modified a hsc question, it first asked to prove |a| = 1 and Re(a) = (1-p)/2, So I decided to omit this and and ask for the whole number since it can be found with those properties I see, thanks for taking the time to write your solution out. ============ $i)...

The only true mechanics was a numerical circular motion connected to 2 fixed points thing. The other 'mechanics' was finding properties of the differential equation, f'' = -k f' - mf where f was x(t) and there were actual numbers there rather than k and m The circle geo was literally a 3U...

Re: HSC 2013 4U Marathon I'm running out of good questions :/

Re: HSC 2013 4U Marathon Yep, the right inequality is kinda similar. ======== $Suppose the polynomial$ \ P(x) = x^3+px^2+px+ 1 \ $has root$ \ - 1 $And complex root$ \ \alpha $Show that$ \alpha = \frac{1-p}{2} + \frac{i}{\sqrt{2}} \sqrt{1+2p-p^2}

Re: 2013 CSSA 4U Maths thoughts? HSC 2009 Q7(b), the only thing different was m replaced with n. EDIT: beeeten

Re: 2013 CSSA 4U Maths thoughts? Just a load of algebra, subbing gradients into that angle between 2 lines formula