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Re: HSC 2013 4U Marathon Yep you are correct, I made a silly error, I will change e and f accordingly.

Re: HSC 2013 4U Marathon Apologies hehe, I'll fix it now.

Re: HSC 2013 4U Marathon Here is another question: $Question Fibonacci - 15 Marks$ $The Fibonacci sequence is a sequence of natural numbers based on the following recurrence relation$ \\ \\ $Where $F_n$denotes the nth Fibonacci number$ \ \\ \\ \ F_0=1 \ \ \ F_1=1 \\ F_n=F_{n-1}+F_{n-2} \ \...

Re: HSC 2013 4U Marathon I have written up a solution, but I will do what you are doing and wait until night and post my solution after yours.

Re: HSC 2013 4U Marathon $Define the nth Triangular number as$ \ \ T_n=\sum_{k=1}^{n}k = 1+2+3+...+n $i) Show that $ \ \ \sum_{k=1}^{\infty} \frac{1}{T_k} = 2\sum_{k=1}^{\infty} \frac{1}{k}-\frac{1}{k+1} \ \ \ \ \ \fbox{2} $ii) Hence evaluate $ \ \ \sum_{k=1}^{\infty} \frac{1}{T_k} \ \ \ \...

Re: HSC 2013 4U Marathon Nice work. I'll try make a question with triangular number properties or something considering how much you like them heh.

Re: HSC 2013 4U Marathon Correct

Re: HSC 2013 4U Marathon $What is the sum of all the natural numbers below 1000 that are multiples of 3 or 5? $

I'm pretty sure it was a joke lol That is the very first technique you learn when doing 3U related rates (Physical Applications of Calculus).

Woah, 4U already? Why are you even learning implicit differentiation o_o Anyway, if we draw our diagram we immediately notice a right angle triangle. Hence by Pythagoras: x^2+y^2=8^2 \ \ \ x^2+y^2=64 Let us differentiate implicitly: x^2+y^2=64 \ \ \ 2x+2y y'=0 -yy'=x \ \ \ \therefore...

There are 3 possible points D can take, remember that.

Ok, I have barely any problems with calculation and problem solving in Physics and Chemistry, however as I have heard the sciences mark really hard the long responses, docking marks for tiny reasons, and it is that reason why it is difficult to get as high. So I have a couple of questions...

Well that just fell apart lol. Thanks

So we all know that irrational numbers can be expressed as an infinite construction of integers in various forms. I.e: 1+\frac{1}{1+\frac{1}{1+1\frac{1}{1+...}}}= \phi e=\sum_{n=0}^{\infty} \frac{1}{n!} \frac{\pi^2}{6}=\sum_{n=1}^{\infty} \frac{1}{n^2} and so on. Is it possible to...

Re: HSC 2013 4U Marathon $Show that$ \ \ \ \ \ \ \ 3=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6....}}}}} $Similarly, evaluate$ \\ \\ 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}}

Great work Realise, really good questions there. (I might try and make a 3U trial)

Re: HSC 2013 4U Marathon Correct

Re: HSC 2013 4U Marathon Close but not quite

Re: HSC 2013 4U Marathon Nice work on the above question, but for this, we can arrive at two variables: \tan^{-1} \frac{y}{x} + \tan^{-1} \frac{x-y}{x+y} = \frac{\pi}{4} Apply the tangent function to everything: y/x + \frac{x-y}{x+y} = 1- \frac{y(x-y)}{x+y} Simplify and we get: 0=0...

Re: HSC 2013 4U Marathon I think i figured out how to justify it hehe (its pretty simple, not sure why I couldn't see it earlier) $Consider$ \ \ \ \langle z, z \rangle - \langle tw, z \rangle = z_1 \bar{z_1} - tw_1 \bar{z_1} + z_2 \bar{z_2}- tw_2 \bar{z_2}+...+z_n \bar{z_n}-tw_n \bar{z_n} \\...