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No, you don’t have to be first, though I imagine it depends on the school and the subject. If your internal competition is strong, it doesn’t matter if a few people outrank you as your internal mark will still be high. I know of people, myself among them who have achieved state ranks when...

99.95 and state ranking student here - these are my 2 cents: 1) Were you guys always top in your cohort (this doesn’t really apply to Top 25 SS students I guess, since rank 5-10’s also get 99.95)? By always, I mean Year 10,11 and 12? Usually first or second in the cohort, top 5 in most...

Every little thing is gonna be alright! http://vimeo.com/m/21681429

First up, let me just say that English Extension 2 is awesome! Congratulations for choosing it. :) Most of the time when people start, they have absolutely no idea what they want to do. Either that or they're already really passionate about something. To be honest, it doesn't matter, as long as...

For the radius question, I proved and used the 'extended' Sine Rule : a/sina = b/sinb = c/sin c = 2R. Works like a dream. :)

Re: HSC 2013 4U Marathon \theta=\pi/2\text{ yields }\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots \text{We know from partial fractions that } \frac{1}{(4n-2)(4n-1)4n}-\frac{1}{4n(4n+1)(4n+2)} = \frac{1}{4}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)+\frac{1}{4n+1}-\frac{1}{4n-1}...

Re: HSC 2013 4U Marathon Isn't this not quite right? It's not specified that the common ratio must be an integer. Here we're missing out on solutions like (4,6,9), (25,30,36), (27,36,48), (44,66,99)...In fact, there are nearly twice as many solutions (I think) if we allow non-integer ratios...

Re: HSC 2013 4U Marathon \frac{s_n}{s_{n-1}}=\binom{n}{n}\prod_{k=0}^{n-1}\frac{\binom{n}{k}}{\binom{n-1}{k}}=\prod_{k=0}^{n-1}\frac{n}{n-k}=\frac{n^n}{n!} \therefore \frac{\left(\frac{s_{n+1}}{s_n} \right)}{\left(\frac{s_{n}}{s_{n-1}} \right)}=\frac{\left(\frac{(n+1)^{n+1}}{(n+1)!}...

Re: HSC 2013 4U Marathon \text{Given that } \binom{n}{r}=0 $ for $ r>n $ or $ r<0, $Show that the sequence defined by$ a_{n}=\sum_{k\ge0}\binom{k}{n-k} $is the sequence of Fibonnaci numbers.$

Re: HSC 2013 4U Marathon Yep - just missing the 'r' from the RHS of your identity in the last two lines. Final answer should be nr/m.

Re: HSC 2013 4U Marathon $Ok, so by the average number of balls, I'm referring to the expected number of balls: the weighted average of all possible values.$ $That is,$ $E(x)=\sum_{k=0}^r kP(k) \text{Interesting to think about what m would be in your formulation.} \text{Clearly }...

Re: HSC 2013 4U Marathon i) is correct. ii) is the correct approach, and very nearly the correct answer - I suspect some minor error somewhere along the line. The numerator should be (n+1)(r+1)=nr+n+r+1

Re: HSC 2013 4U Marathon There are M marbles in a jar, R of which are red. I draw out N marbles without replacing any. i) What is the probability of me drawing out K red marbles? ii) What is the most likely number of red marbles I will draw out? iii) What is the average number of red...

Re: HSC 2013 4U Marathon $ \ \ \sum_{n=1}^{\infty}\frac{1}{n}\sum_{m=1}^{\infty}\\frac{1}{m}\frac{1}{m+n+2}$ $ \ \ \sum_{n=1}^{\infty}\frac{1}{n(n+2)}\sum_{m=1}^{ \infty }\frac{1}{m}-\frac{1}{m+n+2}$ $ \ \...

Re: HSC 2013 4U Marathon $b^2<b^2+b+1<b^2+2b+1=(b+1)^2$ No squares between consecutive squares.

Re: HSC 2013 4U Marathon $z_{k}=a_{k}+ib_{k} $\text{so the inequality says}$ $\sum_{k=1}^{n}|z_{k}|\ge\left|\sum_{k=1}^{n}z_{k}\right|$ $n=2\text{ is the triangle inequality and the induction is straightforward}$...

Re: HSC 2013 4U Marathon This one's the triangle inequality right? Just need a little bit of induction to extend it.

Re: HSC 2013 4U Marathon Intermediate steps included: 11/8 + Sum r=3 to infinite: 1/(r+1)*(1/((r-1)(r+1))+1/(r(r+2))+1/((r+1)(r+3))+....) Which equals 11/8 + 0.5*Sum r=3 to infinite: 1/(r+1)*(1/(r-1)+1/r) Sorry no Latex - still working on my skills in that department.

Re: HSC 2013 4U Marathon Part i: first step change it to 11/8 + 0.5*Sum from r=3 to infinite of (2r-1)/(r^3-r). This comes from 'changing the perspective' from summing sums of 1/k terms to sums of 1/((k+2)k) terms. Then use partial fractions and a couple of telescopes to get 11/8+5/12-1/24=7/4...

Re: HSC 2013 4U Marathon Is it $p+q\le1$? $x^2+y^2\ge2xy$ $x^2(\frac{1-p}{q})+y^2(\frac{1-q}{p})\ge2xy$ $x^2(1-p)p+y^2(1-q)q\ge2pqxy$ $px^2+qy^2\ge p^2x^2+q^2y^2+2pqxy$ $px^2+qy^2\ge (px+qy)^2$